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GRADE 8 PRACTICE

Physics Problems: Thermal phenomena, Grade 8

MCT, temperature, internal energy and heat transfer.

36 problems across 8 sub-topics
Practise in the interactive mode

Here the conditions and analysis of the solution can be read without an account. In the simulator, you need to collect the same analyzes yourself: “Given,” formulas, order of steps and answer.

Problem statements

The answer options are there to check against, and the walkthrough opens under each problem.

36 problems
Position in this set: 1 from 36

Problem #1

Kinetic particle theory and surface phenomena

What concept does the description correspond to: “Liquid rises or falls in a narrow tube because of surface forces and its wetting interaction with the walls.»?

Answer options

  1. \text{Capillary action}
  2. \text{Wetting and non-wetting}
  3. \text{The main ideas of kinetic particle theory}
  4. \text{Thermal equilibrium}
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Given
\text{Description}
\text{Liquid rises or falls in a narrow tube because of surface forces and its wetting interaction with the walls.}
Find

\text{name concepts}

  1. \text{Let's compare approval with definition}

    The narrower a capillary wetted by the liquid, the greater the rise. This effect helps water move through pores in soil and plants.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Capillary action}

Answer\text{Concept} = \text{Capillary action}

Position in this set: 2 from 36

Problem #2

Kinetic particle theory and surface phenomena

What concept does the description correspond to: “Wetting - spreading liquids by hard surfaces; at non-wetting liquid stronger going B drops.»?

Answer options

  1. \text{Wetting and non-wetting}
  2. \text{Capillary action}
  3. \text{The main ideas of kinetic particle theory}
  4. \text{Thermal equilibrium}
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Given
\text{Description}
\text{Wetting - spreading liquids by hard surfaces; at non-wetting liquid stronger going B drops.}
Find

\text{name concepts}

  1. \text{Let's compare approval with definition}

    The shape of a drop or meniscus depends on interactions at the interfaces between the liquid, solid and surrounding gas, as well as the condition of the surface. Water wets clean glass well and wax less well. Comparing just two attractive forces is not enough for a general explanation.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Wetting and non-wetting}

Answer\text{Concept} = \text{Wetting and non-wetting}

Position in this set: 3 from 36

Problem #3

Surface phenomena

Water wets clean glass. How will the height of its rise change if the capillary is replaced with a narrower one?

Answer options

  1. \text{Height will increase}
  2. \text{Height will decrease}
  3. \text{level necessarily will go down Below external}
  4. \text{Diameter not influences on phenomenon}
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Given
\text{liquid}
\text{wets glass}
Find

\text{change heights}

  1. \text{Superficial strength valid by border tubes}
  2. \text{B more narrow tube her action more noticeable relatively weight pillar}h\uparrow

Answer\text{height will increase}

Position in this set: 4 from 36

Problem #4

Temperature and internal energy

What concept does the description correspond to: “The state in which objects in contact have equal temperatures and net heat transfer stops.»?

Answer options

  1. \text{Thermal equilibrium}
  2. \text{Temperature scales}
  3. \text{Internal energy}
  4. \text{Capillary action}
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Given
\text{Description}
\text{The state in which objects in contact have equal temperatures and net heat transfer stops.}
Find

\text{name concepts}

  1. \text{Let's compare approval with definition}

    A thermometer gives the correct temperature after reaching thermal equilibrium with the object being measured.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Thermal equilibrium}

Answer\text{Concept} = \text{Thermal equilibrium}

Position in this set: 5 from 36

Problem #5

Temperature and internal energy

What concept does the description correspond to: “The Celsius and Kelvin scales have equal-sized intervals but different zero points.»?

Answer options

  1. \text{Temperature scales}
  2. \text{Thermal equilibrium}
  3. \text{Internal energy}
  4. \text{Capillary action}
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Given
\text{Description}
\text{The Celsius and Kelvin scales have equal-sized intervals but different zero points.}
Find

\text{name concepts}

  1. \text{Let's compare approval with definition}

    Zero degrees Celsius corresponds to about 273,15 kelvins, and absolute temperature cannot be negative.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Temperature scales}

Answer\text{Concept} = \text{Temperature scales}

Position in this set: 6 from 36

Problem #6

Temperature

The air temperature is 27 °C. Find the absolute temperature, rounding the result to the nearest whole kelvin.

Answer options

  1. 300\,\text{К}
  2. 246\,\text{К}
  3. 273\,\text{К}
  4. 327\,\text{К}
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Given
t
27{}^\circ\text{C}
Find

T

  1. T=t+273{,}15T=27+273{,}15\,\text{К}T = 300{,}15\,\text{К}
  2. T\approx300\,\text{К}

Answer300\,\text{К}

Position in this set: 7 from 36

Problem #7

Thermal conduction, convection and radiation

What concept does the description correspond to: “Transfer of internal energy through particle contact without bulk movement of the substance.»?

Answer options

  1. \text{Thermal conduction}
  2. \text{Convection}
  3. \text{Thermal radiation}
  4. \text{Capillary action}
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Given
\text{Description}
\text{Transfer of internal energy through particle contact without bulk movement of the substance.}
Find

\text{name concepts}

  1. \text{Let's compare approval with definition}

    Metals usually conduct heat well, and air and porous materials are used as heat insulators.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Thermal conduction}

Answer\text{Concept} = \text{Thermal conduction}

Position in this set: 8 from 36

Problem #8

Thermal conduction, convection and radiation

What concept does the description correspond to: “Transfer of internal energy by flowing liquid or gas.»?

Answer options

  1. \text{Convection}
  2. \text{Thermal conduction}
  3. \text{Thermal radiation}
  4. \text{Capillary action}
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Given
\text{Description}
\text{Transfer of internal energy by flowing liquid or gas.}
Find

\text{name concepts}

  1. \text{Let's compare approval with definition}

    In natural convection, motion occurs due to density differences in the gravitational field. Forced convection is created by a pump or fan.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Convection}

Answer\text{Concept} = \text{Convection}

Position in this set: 9 from 36

Problem #9

Heat transfer

Which combination of solutions most strongly reduces all three methods of heat transfer in a thermos?

Answer options

  1. \text{Vacuum gap and reflective walls}
  2. \text{Fat copper wall}
  3. \text{Open wide neck}
  4. \text{Dark matte external surface}
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Given
\text{GOAL}
\text{reduce losses heat}
Find

\text{design}

  1. \text{Vacuum suppresses thermal conductivity and convection}
  2. \text{Reflective surface reduces radiation}\text{Both solutions together}

Answer\text{vacuum gap and reflective walls}

Position in this set: 10 from 36

Problem #10

Heat and thermal energy balance

What concept does the description correspond to: “Energy transferred to or from an object during heat transfer.»?

Answer options

  1. \text{Heat}
  2. \text{Calorimeter and thermal energy balance}
  3. \text{Fuel energy}
  4. \text{Capillary action}
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Given
\text{Description}
\text{Energy transferred to or from an object during heat transfer.}
Find

\text{name concepts}

  1. \text{Let's compare approval with definition}

    During warming without a phase change, heat depends on mass, the substance's specific heat capacity and the temperature change.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Heat}

Answer\text{Concept} = \text{Heat}

Position in this set: 11 from 36

Problem #11

Heat and thermal energy balance

What concept does the description correspond to: “A calorimeter reduces heat exchange with the surroundings. In a thermally insulated system with no external work, heat given out by some objects equals heat received by the others.»?

Answer options

  1. \text{Calorimeter and thermal energy balance}
  2. \text{Heat}
  3. \text{Fuel energy}
  4. \text{Capillary action}
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Given
\text{Description}
\text{A calorimeter reduces heat exchange with the surroundings. In a thermally insulated system with no external work, heat given out by some objects equals heat received by the others.}
Find

\text{name concepts}

  1. \text{Let's compare approval with definition}

    To determine specific heat capacity, measure masses and temperatures, then write an energy balance using one consistent sign convention.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Calorimeter and thermal energy balance}

Answer\text{Concept} = \text{Calorimeter and thermal energy balance}

Position in this set: 12 from 36

Problem #12

Heat

Two kilograms of water were heated by 10 K. The specific heat capacity of water is 4200 J/(kg K). How much heat did the water receive?

Answer options

  1. 84\,\text{кДж}
  2. 21\,\text{кДж}
  3. 42\,\text{кДж}
  4. 840\,\text{кДж}
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Given
m
2\,\text{кг}
c
4200\,\text{Дж/(кг}\cdot\text{К)}
\Delta T
10\,\text{К}
Find

Q

  1. Q=cm\Delta TQ=4200\cdot2\cdot10\,\text{Дж}Q = 84000\,\text{Дж}
  2. Q=84\,\text{кДж}

Answer84\,\text{кДж}

Position in this set: 13 from 36

Problem #73

Kinetic particle theory and surface phenomena

Coefficient superficial tension liquids 0{,}03 N/m. Length borders one surfaces 0{,}2 m. Find force superficial tension by formula F = σl.

Answer options

  1. 0{,}006\,\text{Н}
  2. 0{,}012\,\text{Н}
  3. 0{,}003\,\text{Н}
  4. 0{,}018\,\text{Н}
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Given
r
0{,}03\,\text{Н/м}
l
0{,}2\,\text{м}
Find

F

  1. F=\sigma lF=0{,}03 \cdot 0{,}2F=0{,}006\,\text{Н}

    The force acts along the boundary of the liquid surface.

Answer0{,}006\,\text{Н}

Position in this set: 14 from 36

Problem #74

Kinetic particle theory and surface phenomena

Coefficient superficial tension liquids 0{,}04 N/m. Length borders one surfaces 0{,}3 m. Find force superficial tension by formula F = σl.

Answer options

  1. 0{,}024\,\text{Н}
  2. 0{,}012\,\text{Н}
  3. 0{,}006\,\text{Н}
  4. 0{,}036\,\text{Н}
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Given
r
0{,}04\,\text{Н/м}
l
0{,}3\,\text{м}
Find

F

  1. F=\sigma lF=0{,}04 \cdot 0{,}3F=0{,}012\,\text{Н}

    The force acts along the boundary of the liquid surface.

Answer0{,}012\,\text{Н}

Position in this set: 15 from 36

Problem #75

Kinetic particle theory and surface phenomena

Soapy film pulls mobile crossbar length 10 cm with by force 0{,}006 N. Find coefficient superficial tension. У films two surfaces, therefore F = 2σl.

Answer options

  1. 0{,}06\,\text{Н/м}
  2. 0{,}03\,\text{Н/м}
  3. 0{,}015\,\text{Н/м}
  4. 0{,}09\,\text{Н/м}
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Given
l
10\,\text{см}
F
0{,}006\,\text{Н}
Find

\sigma

  1. a=\frac{l}{100}a=\frac{10}{100}a=0{,}1\,\text{м}

    We give the length of the crossbar to meters.

  2. \sigma=\frac{F}{2a}\sigma=\frac{0{,}006}{(2 \cdot 0{,}1)}\sigma=0{,}03\,\text{Н/м}

    We take into account both sides of the film.

Answer0{,}03\,\text{Н/м}

Position in this set: 16 from 36

Problem #76

Kinetic particle theory and surface phenomena

Soapy film pulls mobile crossbar length 20 cm with by force 0{,}016 N. Find coefficient superficial tension. У films two surfaces, therefore F = 2σl.

Answer options

  1. 0{,}08\,\text{Н/м}
  2. 0{,}02\,\text{Н/м}
  3. 0{,}04\,\text{Н/м}
  4. 0{,}12\,\text{Н/м}
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Given
l
20\,\text{см}
F
0{,}016\,\text{Н}
Find

\sigma

  1. a=\frac{l}{100}a=\frac{20}{100}a=0{,}2\,\text{м}

    We give the length of the crossbar to meters.

  2. \sigma=\frac{F}{2a}\sigma=\frac{0{,}016}{(2 \cdot 0{,}2)}\sigma=0{,}04\,\text{Н/м}

    We take into account both sides of the film.

Answer0{,}04\,\text{Н/м}

Position in this set: 17 from 36

Problem #77

Kinetic particle theory and surface phenomena

Crossbar length 10 cm moved on 5 cm, increasing area soapy films. Find work against strength superficial tension. σ = 0{,}03 N/m; use ΔA = 2lx for two surfaces and W = σΔA.

Answer options

  1. 0{,}0006\,\text{Дж}
  2. 0{,}00015\,\text{Дж}
  3. 0{,}0003\,\text{Дж}
  4. 0{,}0009\,\text{Дж}
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Given
l
10\,\text{см}
x
5\,\text{см}
r
0{,}03\,\text{Н/м}
Find

W

  1. a=\frac{l}{100}a=\frac{10}{100}a=0{,}1\,\text{м}

    We convert the length to meters.

  2. b=\frac{x}{100}b=\frac{5}{100}b=0{,}05\,\text{м}

    We reduce the displacement to meters.

  3. \Delta A=2ab\Delta A=2 \cdot 0{,}1 \cdot 0{,}05\Delta A=0{,}01\,\text{м}^{2}

    We consider the increase in the total area of both surfaces.

  4. W=\sigma\Delta AW=0{,}03 \cdot 0{,}01W=0{,}0003\,\text{Дж}

    Work increases surface energy.

Answer0{,}0003\,\text{Дж}

Position in this set: 18 from 36

Problem #78

Kinetic particle theory and surface phenomena

Crossbar length 20 cm moved on 10 cm, increasing area soapy films. Find work against strength superficial tension. σ = 0{,}04 N/m; use ΔA = 2lx for two surfaces and W = σΔA.

Answer options

  1. 0{,}0032\,\text{Дж}
  2. 0{,}0008\,\text{Дж}
  3. 0{,}0048\,\text{Дж}
  4. 0{,}0016\,\text{Дж}
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Given
l
20\,\text{см}
x
10\,\text{см}
r
0{,}04\,\text{Н/м}
Find

W

  1. a=\frac{l}{100}a=\frac{20}{100}a=0{,}2\,\text{м}

    We convert the length to meters.

  2. b=\frac{x}{100}b=\frac{10}{100}b=0{,}1\,\text{м}

    We reduce the displacement to meters.

  3. \Delta A=2ab\Delta A=2 \cdot 0{,}2 \cdot 0{,}1\Delta A=0{,}04\,\text{м}^{2}

    We consider the increase in the total area of both surfaces.

  4. W=\sigma\Delta AW=0{,}04 \cdot 0{,}04W=0{,}0016\,\text{Дж}

    Work increases surface energy.

Answer0{,}0016\,\text{Дж}

Position in this set: 19 from 36

Problem #79

Temperature and internal energy

When heating temperature body grew up from 18 up to 48 °C. Find Temperature change.

Answer options

  1. 60\,\text{°C}
  2. 30\,\text{°C}
  3. 15\,\text{°C}
  4. 90\,\text{°C}
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Given
T_1
18\,\text{°C}
T_2
48\,\text{°C}
Find

\Delta T

  1. \Delta T=T_2-T_1\Delta T=48 - 18\Delta T=30\,\text{°C}

    Find the difference between the final and initial temperatures.

Answer30\,\text{°C}

Position in this set: 20 from 36

Problem #80

Temperature and internal energy

When heating temperature body grew up from 23 up to 63 °C. Find Temperature change.

Answer options

  1. 80\,\text{°C}
  2. 20\,\text{°C}
  3. 40\,\text{°C}
  4. 120\,\text{°C}
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Given
T_1
23\,\text{°C}
T_2
63\,\text{°C}
Find

\Delta T

  1. \Delta T=T_2-T_1\Delta T=63 - 23\Delta T=40\,\text{°C}

    Find the difference between the final and initial temperatures.

Answer40\,\text{°C}

Position in this set: 21 from 36

Problem #81

Temperature and internal energy

For promotion temperature sample on 1 °C Required 200 J. him handed over 4000 J warmth without losses. Initial temperature 20 °C. Find final temperature; phase transitions no.

Answer options

  1. 80\,\text{°C}
  2. 20\,\text{°C}
  3. 40\,\text{°C}
  4. 120\,\text{°C}
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Given
C
200\,\text{Дж/°C}
Q
4000\,\text{Дж}
T_0
20\,\text{°C}
Find

T

  1. \Delta T=\frac{Q}{C}\Delta T=\frac{4000}{200}\Delta T=20\,\text{°C}

    We use the heat capacity given in the condition.

  2. T=T_0+\Delta TT=20 + 20T=40\,\text{°C}

    We add the increase to the initial temperature.

Answer40\,\text{°C}

Position in this set: 22 from 36

Problem #82

Temperature and internal energy

For promotion temperature sample on 1 °C Required 300 J. him handed over 9000 J warmth without losses. Initial temperature 15 °C. Find final temperature; phase transitions no.

Answer options

  1. 90\,\text{°C}
  2. 22{,}5\,\text{°C}
  3. 135\,\text{°C}
  4. 45\,\text{°C}
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Given
C
300\,\text{Дж/°C}
Q
9000\,\text{Дж}
T_0
15\,\text{°C}
Find

T

  1. \Delta T=\frac{Q}{C}\Delta T=\frac{9000}{300}\Delta T=30\,\text{°C}

    We use the heat capacity given in the condition.

  2. T=T_0+\Delta TT=15 + 30T=45\,\text{°C}

    We add the increase to the initial temperature.

Answer45\,\text{°C}

Position in this set: 23 from 36

Problem #83

Temperature and internal energy

sample received 10000 J from heater and gave surrounding environment 2000 J. For heating on 1 °C Required 400 J. Initial temperature 18 °C. Find final temperature. Job body not commits, phase transitions no.

Answer options

  1. 76\,\text{°C}
  2. 19\,\text{°C}
  3. 114\,\text{°C}
  4. 38\,\text{°C}
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Given
Q_1
10000\,\text{Дж}
Q_2
2000\,\text{Дж}
C
400\,\text{Дж/°C}
T_0
18\,\text{°C}
Find

T

  1. Q=Q_1-Q_2Q=10000 - 2000Q=8000\,\text{Дж}

    We take into account both directions of heat transfer.

  2. \Delta T=\frac{Q}{C}\Delta T=\frac{8000}{400}\Delta T=20\,\text{°C}

    Only the remaining energy increases the temperature.

  3. T=T_0+\Delta TT=18 + 20T=38\,\text{°C}

    Finding the final temperature.

Answer38\,\text{°C}

Position in this set: 24 from 36

Problem #84

Temperature and internal energy

sample received 15000 J from heater and gave surrounding environment 3000 J. For heating on 1 °C Required 600 J. Initial temperature 22 °C. Find final temperature. Job body not commits, phase transitions no.

Answer options

  1. 42\,\text{°C}
  2. 84\,\text{°C}
  3. 21\,\text{°C}
  4. 126\,\text{°C}
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Given
Q_1
15000\,\text{Дж}
Q_2
3000\,\text{Дж}
C
600\,\text{Дж/°C}
T_0
22\,\text{°C}
Find

T

  1. Q=Q_1-Q_2Q=15000 - 3000Q=12000\,\text{Дж}

    We take into account both directions of heat transfer.

  2. \Delta T=\frac{Q}{C}\Delta T=\frac{12000}{600}\Delta T=20\,\text{°C}

    Only the remaining energy increases the temperature.

  3. T=T_0+\Delta TT=22 + 20T=42\,\text{°C}

    Finding the final temperature.

Answer42\,\text{°C}

Position in this set: 25 from 36

Problem #85

Thermal conduction, convection and radiation

When heating temperature body grew up from 18 up to 48 °C. Find Temperature change.

Answer options

  1. 60\,\text{°C}
  2. 15\,\text{°C}
  3. 30\,\text{°C}
  4. 90\,\text{°C}
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Given
T_1
18\,\text{°C}
T_2
48\,\text{°C}
Find

\Delta T

  1. \Delta T=T_2-T_1\Delta T=48 - 18\Delta T=30\,\text{°C}

    Find the difference between the final and initial temperatures.

Answer30\,\text{°C}

Position in this set: 26 from 36

Problem #86

Thermal conduction, convection and radiation

When heating temperature body grew up from 23 up to 63 °C. Find Temperature change.

Answer options

  1. 80\,\text{°C}
  2. 20\,\text{°C}
  3. 120\,\text{°C}
  4. 40\,\text{°C}
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Given
T_1
23\,\text{°C}
T_2
63\,\text{°C}
Find

\Delta T

  1. \Delta T=T_2-T_1\Delta T=63 - 23\Delta T=40\,\text{°C}

    Find the difference between the final and initial temperatures.

Answer40\,\text{°C}

Position in this set: 27 from 36

Problem #87

Thermal conduction, convection and radiation

For promotion temperature sample on 1 °C Required 200 J. him handed over 4000 J warmth without losses. Initial temperature 20 °C. Find final temperature; phase transitions no.

Answer options

  1. 80\,\text{°C}
  2. 20\,\text{°C}
  3. 120\,\text{°C}
  4. 40\,\text{°C}
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Given
C
200\,\text{Дж/°C}
Q
4000\,\text{Дж}
T_0
20\,\text{°C}
Find

T

  1. \Delta T=\frac{Q}{C}\Delta T=\frac{4000}{200}\Delta T=20\,\text{°C}

    We use the heat capacity given in the condition.

  2. T=T_0+\Delta TT=20 + 20T=40\,\text{°C}

    We add the increase to the initial temperature.

Answer40\,\text{°C}

Position in this set: 28 from 36

Problem #88

Thermal conduction, convection and radiation

For promotion temperature sample on 1 °C Required 300 J. him handed over 9000 J warmth without losses. Initial temperature 15 °C. Find final temperature; phase transitions no.

Answer options

  1. 45\,\text{°C}
  2. 90\,\text{°C}
  3. 22{,}5\,\text{°C}
  4. 135\,\text{°C}
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Given
C
300\,\text{Дж/°C}
Q
9000\,\text{Дж}
T_0
15\,\text{°C}
Find

T

  1. \Delta T=\frac{Q}{C}\Delta T=\frac{9000}{300}\Delta T=30\,\text{°C}

    We use the heat capacity given in the condition.

  2. T=T_0+\Delta TT=15 + 30T=45\,\text{°C}

    We add the increase to the initial temperature.

Answer45\,\text{°C}

Position in this set: 29 from 36

Problem #89

Thermal conduction, convection and radiation

sample received 10000 J from heater and gave surrounding environment 2000 J. For heating on 1 °C Required 400 J. Initial temperature 18 °C. Find final temperature. Job body not commits, phase transitions no.

Answer options

  1. 38\,\text{°C}
  2. 76\,\text{°C}
  3. 19\,\text{°C}
  4. 114\,\text{°C}
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Given
Q_1
10000\,\text{Дж}
Q_2
2000\,\text{Дж}
C
400\,\text{Дж/°C}
T_0
18\,\text{°C}
Find

T

  1. Q=Q_1-Q_2Q=10000 - 2000Q=8000\,\text{Дж}

    We take into account both directions of heat transfer.

  2. \Delta T=\frac{Q}{C}\Delta T=\frac{8000}{400}\Delta T=20\,\text{°C}

    Only the remaining energy increases the temperature.

  3. T=T_0+\Delta TT=18 + 20T=38\,\text{°C}

    Finding the final temperature.

Answer38\,\text{°C}

Position in this set: 30 from 36

Problem #90

Thermal conduction, convection and radiation

sample received 15000 J from heater and gave surrounding environment 3000 J. For heating on 1 °C Required 600 J. Initial temperature 22 °C. Find final temperature. Job body not commits, phase transitions no.

Answer options

  1. 84\,\text{°C}
  2. 42\,\text{°C}
  3. 21\,\text{°C}
  4. 126\,\text{°C}
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Given
Q_1
15000\,\text{Дж}
Q_2
3000\,\text{Дж}
C
600\,\text{Дж/°C}
T_0
22\,\text{°C}
Find

T

  1. Q=Q_1-Q_2Q=15000 - 3000Q=12000\,\text{Дж}

    We take into account both directions of heat transfer.

  2. \Delta T=\frac{Q}{C}\Delta T=\frac{12000}{600}\Delta T=20\,\text{°C}

    Only the remaining energy increases the temperature.

  3. T=T_0+\Delta TT=22 + 20T=42\,\text{°C}

    Finding the final temperature.

Answer42\,\text{°C}

Position in this set: 31 from 36

Problem #91

Heat and thermal energy balance

water mass 2 kg heated up on 10 °C. Specific heat capacity water 4200 J/(kg °C). Find Heat.

Answer options

  1. 168000\,\text{Дж}
  2. 42000\,\text{Дж}
  3. 252000\,\text{Дж}
  4. 84000\,\text{Дж}
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Given
m
2\,\text{кг}
d
10\,\text{°C}
c
4200\,\text{Дж/(кг·°C)}
Find

Q

  1. Q=cm\Delta TQ=4200 \cdot 2 \cdot 10Q=84000\,\text{Дж}

    We use temperature change.

Answer84000\,\text{Дж}

Position in this set: 32 from 36

Problem #92

Heat and thermal energy balance

water mass 3 kg heated up on 20 °C. Specific heat capacity water 4200 J/(kg °C). Find Heat.

Answer options

  1. 252000\,\text{Дж}
  2. 504000\,\text{Дж}
  3. 126000\,\text{Дж}
  4. 756000\,\text{Дж}
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Given
m
3\,\text{кг}
d
20\,\text{°C}
c
4200\,\text{Дж/(кг·°C)}
Find

Q

  1. Q=cm\Delta TQ=4200 \cdot 3 \cdot 20Q=252000\,\text{Дж}

    We use temperature change.

Answer252000\,\text{Дж}

Position in this set: 33 from 36

Problem #93

Heat and thermal energy balance

On heating substances from 20 up to 60 °C spent 80000 J. Specific heat capacity 1000 J/(kg °C). Find mass, neglecting losses.

Answer options

  1. 2\,\text{кг}
  2. 4\,\text{кг}
  3. 1\,\text{кг}
  4. 6\,\text{кг}
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Given
T_1
20\,\text{°C}
T_2
60\,\text{°C}
Q
80000\,\text{Дж}
c
1000\,\text{Дж/(кг·°C)}
Find

m

  1. \Delta T=T_2-T_1\Delta T=60 - 20\Delta T=40\,\text{°C}

    Finding the temperature range.

  2. m=\frac{Q}{c\Delta T}m=\frac{80000}{(1000 \cdot 40)}m=2\,\text{кг}

    We express the mass from the heating equation.

Answer2\,\text{кг}

Position in this set: 34 from 36

Problem #94

Heat and thermal energy balance

On heating substances from 10 up to 50 °C spent 144000 J. Specific heat capacity 900 J/(kg °C). Find mass, neglecting losses.

Answer options

  1. 8\,\text{кг}
  2. 4\,\text{кг}
  3. 2\,\text{кг}
  4. 12\,\text{кг}
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Given
T_1
10\,\text{°C}
T_2
50\,\text{°C}
Q
144000\,\text{Дж}
c
900\,\text{Дж/(кг·°C)}
Find

m

  1. \Delta T=T_2-T_1\Delta T=50 - 10\Delta T=40\,\text{°C}

    Finding the temperature range.

  2. m=\frac{Q}{c\Delta T}m=\frac{144000}{(900 \cdot 40)}m=4\,\text{кг}

    We express the mass from the heating equation.

Answer4\,\text{кг}

Position in this set: 35 from 36

Problem #95

Heat and thermal energy balance

Mixed 1 kg water at 80 °C and 3 kg water at 20 °C. Find established temperature. Heat capacity vessel and losses neglect.

Answer options

  1. 70\,\text{°C}
  2. 35\,\text{°C}
  3. 17{,}5\,\text{°C}
  4. 105\,\text{°C}
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Given
m_1
1\,\text{кг}
T_1
80\,\text{°C}
m_2
3\,\text{кг}
T_2
20\,\text{°C}
Find

T

  1. a=m_1T_1a=1 \cdot 80a=80\,\text{кг·°C}

    Heat balance: cold water receives heat from hot water.

  2. b=m_2T_2b=3 \cdot 20b=60\,\text{кг·°C}

    The specific heat capacity of both portions is the same and decreases.

  3. T=\frac{m_1T_1+m_2T_2}{m_1+m_2}T=\frac{(80 + 60)}{(1 + 3)}T=35\,\text{°C}

    The temperature lies between the initial values.

Answer35\,\text{°C}

Position in this set: 36 from 36

Problem #96

Heat and thermal energy balance

Mixed 2 kg water at 70 °C and 3 kg water at 20 °C. Find established temperature. Heat capacity vessel and losses neglect.

Answer options

  1. 80\,\text{°C}
  2. 20\,\text{°C}
  3. 40\,\text{°C}
  4. 120\,\text{°C}
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Given
m_1
2\,\text{кг}
T_1
70\,\text{°C}
m_2
3\,\text{кг}
T_2
20\,\text{°C}
Find

T

  1. a=m_1T_1a=2 \cdot 70a=140\,\text{кг·°C}

    Heat balance: cold water receives heat from hot water.

  2. b=m_2T_2b=3 \cdot 20b=60\,\text{кг·°C}

    The specific heat capacity of both portions is the same and decreases.

  3. T=\frac{m_1T_1+m_2T_2}{m_1+m_2}T=\frac{(140 + 60)}{(2 + 3)}T=40\,\text{°C}

    The temperature lies between the initial values.

Answer40\,\text{°C}

The theory behind these problems

Work through the topic if a problem will not give in.

Kinetic particle theory and surface phenomenaThe ideas of kinetic particle theory, states of matter, wetting and capillaries.Temperature and internal energyThermal equilibrium, temperature scales and ways to change internal energy.Thermal conduction, convection and radiationThree methods of heat transfer and their uses.Heat and thermal energy balanceSpecific heat capacity, calorimetry, fuel and energy conservation.

Formulas by topic

Look up symbols and units of measurement in the reference.

Height and depthh · мHeatQ · ДжMassm · кгTemperatureT, t · К, °CSpecific heat capacityc · Дж/(кг·К)EnergyE, W · Дж