fizikzauglom

GRADE 8 PRACTICE

Physics Problems: Change in state of aggregation, Grade 8

Melting, evaporation, moisture and boiling.

27 problems across 6 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.

27 problems
Position in this set: 1 from 27

Problem #13

Melting, solidification and graphs

What concept does the description correspond to: “Melting changes a solid into a liquid; solidification is the reverse change.»?

Answer options

  1. \text{Melting and solidification}
  2. \text{Melting point}
  3. \text{Heating graph}
  4. \text{Capillary action}
Show the walkthroughHide the walkthrough
Given
\text{Description}
\text{Melting changes a solid into a liquid; solidification is the reverse change.}
Find

\text{name concepts}

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

    For a pure crystalline substance, both processes at a given pressure occur at the same temperature.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Melting and solidification}

Answer\text{Concept} = \text{Melting and solidification}

Position in this set: 2 from 27

Problem #14

Melting, solidification and graphs

What concept does the description correspond to: “Temperature of phase equilibrium of solid and liquid states at a given pressure.»?

Answer options

  1. \text{Melting point}
  2. \text{Melting and solidification}
  3. \text{Heating graph}
  4. \text{Capillary action}
Show the walkthroughHide the walkthrough
Given
\text{Description}
\text{Temperature of phase equilibrium of solid and liquid states at a given pressure.}
Find

\text{name concepts}

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

    During equilibrium melting of a pure crystalline substance at constant pressure, energy changes particle arrangement and interactions while temperature stays constant. The molecules themselves are not destroyed. Pressure and impurities can change the transition temperature.

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

Answer\text{Concept} = \text{Melting point}

Position in this set: 3 from 27

Problem #15

Phase transitions

How much heat is needed to melt 0.5 kg of ice at 0 °C? The specific heat of fusion is 334 kJ/kg.

Answer options

  1. 167\,\text{кДж}
  2. 83{,}5\,\text{кДж}
  3. 334\,\text{кДж}
  4. 668\,\text{кДж}
Show the walkthroughHide the walkthrough
Given
m
0{,}5\,\text{кг}
\lambda
334\,\text{кДж/кг}
Find

Q

  1. Q=\lambda mQ=334\cdot0{,}5\,\text{кДж}Q = 167\,\text{кДж}

Answer167\,\text{кДж}

Position in this set: 4 from 27

Problem #16

Evaporation, condensation and humidity

What concept does the description correspond to: “Vapour in dynamic equilibrium with a liquid at a given temperature: the average rates of evaporation and condensation are equal.»?

Answer options

  1. \text{Saturated vapour}
  2. \text{Evaporation and condensation}
  3. \text{Relative humidity}
  4. \text{Capillary action}
Show the walkthroughHide the walkthrough
Given
\text{Description}
\text{Vapour in dynamic equilibrium with a liquid at a given temperature: the average rates of evaporation and condensation are equal.}
Find

\text{name concepts}

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

    Molecules continue to cross in both directions. In a closed vessel, saturation is reached if there is enough liquid and some remains; all the liquid may evaporate before that happens. Saturated vapour density depends on temperature. Unsaturated vapour has a lower density than saturated vapour at the same temperature, and evaporation can exceed condensation.

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

Answer\text{Concept} = \text{Saturated vapour}

Position in this set: 5 from 27

Problem #17

Evaporation, condensation and humidity

What concept does the description correspond to: “Evaporation occurs from the surface of the liquid, and condensation turns the vapor back into liquid.»?

Answer options

  1. \text{Evaporation and condensation}
  2. \text{Saturated vapour}
  3. \text{Relative humidity}
  4. \text{Capillary action}
Show the walkthroughHide the walkthrough
Given
\text{Description}
\text{Evaporation occurs from the surface of the liquid, and condensation turns the vapor back into liquid.}
Find

\text{name concepts}

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

    Heating, wind and increased surface area speed up evaporation; the fastest molecules carry energy away and cool the liquid.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Evaporation and condensation}

Answer\text{Concept} = \text{Evaporation and condensation}

Position in this set: 6 from 27

Problem #18

Humidity

The density of water vapor in the room is 8.5 g/m³, and the density of saturated vapor at the same temperature is 17 g/m³. Find the relative humidity.

Answer options

  1. 50\,\%
  2. 25\,\%
  3. 100\,\%
  4. 200\,\%
Show the walkthroughHide the walkthrough
Given
\rho
8{,}5\,\text{г/м}^{3}
\rho_{\text{нас}}
17\,\text{г/м}^{3}
Find

\varphi

  1. \varphi=\frac{\rho}{\rho_{\text{нас}}}\varphi=\frac{8{,}5}{17}\varphi = 0{,}5
  2. \varphi=50\,\%

Answer50\,\%

Position in this set: 7 from 27

Problem #19

Boiling and vaporization

What concept does the description correspond to: “Intense vaporization throughout a liquid, with vapour bubbles forming and growing within it.»?

Answer options

  1. \text{Boiling}
  2. \text{Specific latent heat of vaporization}
  3. \text{Boiling point and pressure}
  4. \text{Capillary action}
Show the walkthroughHide the walkthrough
Given
\text{Description}
\text{Intense vaporization throughout a liquid, with vapour bubbles forming and growing within it.}
Find

\text{name concepts}

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

    Boiling begins when the saturated vapor pressure is equal to the external pressure.

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

Answer\text{Concept} = \text{Boiling}

Position in this set: 8 from 27

Problem #20

Boiling and vaporization

What concept does the description correspond to: “The energy required to convert a unit mass of liquid into vapor at the boiling point.»?

Answer options

  1. \text{Specific latent heat of vaporization}
  2. \text{Boiling}
  3. \text{Boiling point and pressure}
  4. \text{Capillary action}
Show the walkthroughHide the walkthrough
Given
\text{Description}
\text{The energy required to convert a unit mass of liquid into vapor at the boiling point.}
Find

\text{name concepts}

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

    Condensing the same mass of vapour releases the same magnitude of energy.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Specific latent heat of vaporization}

Answer\text{Concept} = \text{Specific latent heat of vaporization}

Position in this set: 9 from 27

Problem #21

Boiling

To convert it into steam, take 0.2 kg of water at boiling point. The specific heat of vaporization is 2.3 MJ/kg. Find the required heat.

Answer options

  1. 460\,\text{кДж}
  2. 46\,\text{кДж}
  3. 1150\,\text{кДж}
  4. 2300\,\text{кДж}
Show the walkthroughHide the walkthrough
Given
m
0{,}2\,\text{кг}
L
2{,}3\,\text{МДж/кг}2300\,\text{кДж/кг}
Find

Q

  1. Q=LmQ=2300\cdot0{,}2\,\text{кДж}Q = 460\,\text{кДж}

Answer460\,\text{кДж}

Position in this set: 10 from 27

Problem #97

Melting, solidification and graphs

Ice mass 0{,}2 kg is located at 0 °C. How much warmth need for him melting? Specific latent heat of fusion 334000 J/kg.

Answer options

  1. 66800\,\text{Дж}
  2. 133600\,\text{Дж}
  3. 33400\,\text{Дж}
  4. 200400\,\text{Дж}
Show the walkthroughHide the walkthrough
Given
m
0{,}2\,\text{кг}
l
334000\,\text{Дж/кг}
Find

Q

  1. Q=\lambda mQ=0{,}2 \cdot 334000Q=66800\,\text{Дж}

    During melting, the temperature of the ice does not increase.

Answer66800\,\text{Дж}

Position in this set: 11 from 27

Problem #98

Melting, solidification and graphs

Ice mass 0{,}5 kg is located at 0 °C. How much warmth need for him melting? Specific latent heat of fusion 334000 J/kg.

Answer options

  1. 334000\,\text{Дж}
  2. 167000\,\text{Дж}
  3. 83500\,\text{Дж}
  4. 501000\,\text{Дж}
Show the walkthroughHide the walkthrough
Given
m
0{,}5\,\text{кг}
l
334000\,\text{Дж/кг}
Find

Q

  1. Q=\lambda mQ=0{,}5 \cdot 334000Q=167000\,\text{Дж}

    During melting, the temperature of the ice does not increase.

Answer167000\,\text{Дж}

Position in this set: 12 from 27

Problem #99

Melting, solidification and graphs

heater conveys ice power 334 W without losses. Mass ice 0{,}3 kg, initial temperature 0 °C, Specific latent heat of fusion 334000 J/kg. Find time full melting.

Answer options

  1. 600\,\text{с}
  2. 300\,\text{с}
  3. 150\,\text{с}
  4. 900\,\text{с}
Show the walkthroughHide the walkthrough
Given
P
334\,\text{Вт}
m
0{,}3\,\text{кг}
l
334000\,\text{Дж/кг}
Find

t

  1. Q=m\lambdaQ=0{,}3 \cdot 334000Q=100200\,\text{Дж}

    Finding the energy for the phase transition.

  2. t=\frac{Q}{P}t=\frac{100200}{334}t=300\,\text{с}

    At constant power, time is equal to energy divided by power.

Answer300\,\text{с}

Position in this set: 13 from 27

Problem #100

Melting, solidification and graphs

heater conveys ice power 668 W without losses. Mass ice 0{,}4 kg, initial temperature 0 °C, Specific latent heat of fusion 334000 J/kg. Find time full melting.

Answer options

  1. 400\,\text{с}
  2. 100\,\text{с}
  3. 200\,\text{с}
  4. 600\,\text{с}
Show the walkthroughHide the walkthrough
Given
P
668\,\text{Вт}
m
0{,}4\,\text{кг}
l
334000\,\text{Дж/кг}
Find

t

  1. Q=m\lambdaQ=0{,}4 \cdot 334000Q=133600\,\text{Дж}

    Finding the energy for the phase transition.

  2. t=\frac{Q}{P}t=\frac{133600}{668}t=200\,\text{с}

    At constant power, time is equal to energy divided by power.

Answer200\,\text{с}

Position in this set: 14 from 27

Problem #101

Melting, solidification and graphs

Ice mass 0{,}2 kg at 0 °C completely melted down, then water heated up up to 20 °C. Find total warmth. Specific latent heat of fusion 334000 J/kg, heat capacity water 4200 J/(kg °C); losses no.

Answer options

  1. 167200\,\text{Дж}
  2. 41800\,\text{Дж}
  3. 83600\,\text{Дж}
  4. 250800\,\text{Дж}
Show the walkthroughHide the walkthrough
Given
m
0{,}2\,\text{кг}
T
20\,\text{°C}
l
334000\,\text{Дж/кг}
c
4200\,\text{Дж/(кг·°C)}
Find

Q

  1. Q_1=m\lambdaQ_1=0{,}2 \cdot 334000Q_1=66800\,\text{Дж}

    First, energy is spent on melting.

  2. Q_2=cmTQ_2=4200 \cdot 0{,}2 \cdot 20Q_2=16800\,\text{Дж}

    The resulting water is then heated.

  3. Q=Q_1+Q_2Q=66800 + 16800Q=83600\,\text{Дж}

    We add up the energies of two different processes.

Answer83600\,\text{Дж}

Position in this set: 15 from 27

Problem #102

Melting, solidification and graphs

Ice mass 0{,}3 kg at 0 °C completely melted down, then water heated up up to 30 °C. Find total warmth. Specific latent heat of fusion 334000 J/kg, heat capacity water 4200 J/(kg °C); losses no.

Answer options

  1. 276000\,\text{Дж}
  2. 69000\,\text{Дж}
  3. 414000\,\text{Дж}
  4. 138000\,\text{Дж}
Show the walkthroughHide the walkthrough
Given
m
0{,}3\,\text{кг}
T
30\,\text{°C}
l
334000\,\text{Дж/кг}
c
4200\,\text{Дж/(кг·°C)}
Find

Q

  1. Q_1=m\lambdaQ_1=0{,}3 \cdot 334000Q_1=100200\,\text{Дж}

    First, energy is spent on melting.

  2. Q_2=cmTQ_2=4200 \cdot 0{,}3 \cdot 30Q_2=37800\,\text{Дж}

    The resulting water is then heated.

  3. Q=Q_1+Q_2Q=100200 + 37800Q=138000\,\text{Дж}

    We add up the energies of two different processes.

Answer138000\,\text{Дж}

Position in this set: 16 from 27

Problem #103

Evaporation, condensation and humidity

Water vapor density 8 g/m³, Saturated vapor density at that same temperature 20 g/m³. Find relative humidity.

Answer options

  1. 80\,\text{\%}
  2. 40\,\text{\%}
  3. 20\,\text{\%}
  4. 120\,\text{\%}
Show the walkthroughHide the walkthrough
Given
r
8\,\text{г/м}^{3}
R
20\,\text{г/м}^{3}
Find

\varphi

  1. \varphi=\frac{\rho}{\rho_{\text{нас}}}\cdot100\varphi=\frac{8}{20} \cdot 100\varphi=40\,\text{\%}

    Both densities refer to the same temperature.

Answer40\,\text{\%}

Position in this set: 17 from 27

Problem #104

Evaporation, condensation and humidity

Water vapor density 12 g/m³, Saturated vapor density at that same temperature 24 g/m³. Find relative humidity.

Answer options

  1. 100\,\text{\%}
  2. 25\,\text{\%}
  3. 50\,\text{\%}
  4. 150\,\text{\%}
Show the walkthroughHide the walkthrough
Given
r
12\,\text{г/м}^{3}
R
24\,\text{г/м}^{3}
Find

\varphi

  1. \varphi=\frac{\rho}{\rho_{\text{нас}}}\cdot100\varphi=\frac{12}{24} \cdot 100\varphi=50\,\text{\%}

    Both densities refer to the same temperature.

Answer50\,\text{\%}

Position in this set: 18 from 27

Problem #105

Evaporation, condensation and humidity

Volume rooms 30 m³, Relative humidity 50 %, Saturated vapor density 20 g/m³. Find mass water couple B room.

Answer options

  1. 600\,\text{г}
  2. 150\,\text{г}
  3. 300\,\text{г}
  4. 900\,\text{г}
Show the walkthroughHide the walkthrough
Given
V
30\,\text{м}^{3}
f
50\,\text{\%}
R
20\,\text{г/м}^{3}
Find

m

  1. \rho=\rho_{\text{нас}}\frac{\varphi}{100}\rho=\frac{20 \cdot 50}{100}\rho=10\,\text{г/м}^{3}

    Find the actual vapor density.

  2. m=\rho Vm=10 \cdot 30m=300\,\text{г}

    Multiply density by volume.

Answer300\,\text{г}

Position in this set: 19 from 27

Problem #106

Evaporation, condensation and humidity

Volume rooms 40 m³, Relative humidity 60 %, Saturated vapor density 25 g/m³. Find mass water couple B room.

Answer options

  1. 1200\,\text{г}
  2. 300\,\text{г}
  3. 1800\,\text{г}
  4. 600\,\text{г}
Show the walkthroughHide the walkthrough
Given
V
40\,\text{м}^{3}
f
60\,\text{\%}
R
25\,\text{г/м}^{3}
Find

m

  1. \rho=\rho_{\text{нас}}\frac{\varphi}{100}\rho=\frac{25 \cdot 60}{100}\rho=15\,\text{г/м}^{3}

    Find the actual vapor density.

  2. m=\rho Vm=15 \cdot 40m=600\,\text{г}

    Multiply density by volume.

Answer600\,\text{г}

Position in this set: 20 from 27

Problem #107

Evaporation, condensation and humidity

B closed room volume 40 m³ original density couple 15 g/m³. After cooling Saturated vapor density became 10 g/m³. Which one mass water will condense? Volume rooms permanent, 15 more 10.

Answer options

  1. 400\,\text{г}
  2. 100\,\text{г}
  3. 600\,\text{г}
  4. 200\,\text{г}
Show the walkthroughHide the walkthrough
Given
V
40\,\text{м}^{3}
r
15\,\text{г/м}^{3}
R
10\,\text{г/м}^{3}
Find

m

  1. m_1=\rho Vm_1=15 \cdot 40m_1=600\,\text{г}

    Find the initial mass of steam.

  2. m_2=\rho_{\text{нас}}Vm_2=10 \cdot 40m_2=400\,\text{г}

    After cooling, no more than the saturation mass may remain in the vapor.

  3. m=m_1-m_2m=600 - 400m=200\,\text{г}

    Excess steam turns into liquid.

Answer200\,\text{г}

Position in this set: 21 from 27

Problem #108

Evaporation, condensation and humidity

B closed room volume 50 m³ original density couple 18 g/m³. After cooling Saturated vapor density became 12 g/m³. Which one mass water will condense? Volume rooms permanent, 18 more 12.

Answer options

  1. 300\,\text{г}
  2. 600\,\text{г}
  3. 150\,\text{г}
  4. 900\,\text{г}
Show the walkthroughHide the walkthrough
Given
V
50\,\text{м}^{3}
r
18\,\text{г/м}^{3}
R
12\,\text{г/м}^{3}
Find

m

  1. m_1=\rho Vm_1=18 \cdot 50m_1=900\,\text{г}

    Find the initial mass of steam.

  2. m_2=\rho_{\text{нас}}Vm_2=12 \cdot 50m_2=600\,\text{г}

    After cooling, no more than the saturation mass may remain in the vapor.

  3. m=m_1-m_2m=900 - 600m=300\,\text{г}

    Excess steam turns into liquid.

Answer300\,\text{г}

Position in this set: 22 from 27

Problem #109

Boiling and vaporization

Water mass 0{,}1 kg already boils. How much warmth need, to turn her B steam? Specific latent heat of vaporization 2260000 J/kg.

Answer options

  1. 452000\,\text{Дж}
  2. 113000\,\text{Дж}
  3. 226000\,\text{Дж}
  4. 678000\,\text{Дж}
Show the walkthroughHide the walkthrough
Given
m
0{,}1\,\text{кг}
L
2260000\,\text{Дж/кг}
Find

Q

  1. Q=mLQ=0{,}1 \cdot 2260000Q=226000\,\text{Дж}

    Energy goes into the phase transition at a constant temperature.

Answer226000\,\text{Дж}

Position in this set: 23 from 27

Problem #110

Boiling and vaporization

Water mass 0{,}2 kg already boils. How much warmth need, to turn her B steam? Specific latent heat of vaporization 2260000 J/kg.

Answer options

  1. 904000\,\text{Дж}
  2. 226000\,\text{Дж}
  3. 1356000\,\text{Дж}
  4. 452000\,\text{Дж}
Show the walkthroughHide the walkthrough
Given
m
0{,}2\,\text{кг}
L
2260000\,\text{Дж/кг}
Find

Q

  1. Q=mLQ=0{,}2 \cdot 2260000Q=452000\,\text{Дж}

    Energy goes into the phase transition at a constant temperature.

Answer452000\,\text{Дж}

Position in this set: 24 from 27

Problem #111

Boiling and vaporization

Boiling water transmit power 1130 W B current 200 s. Which one mass will evaporate at absence losses? Specific latent heat of vaporization 2260000 J/kg.

Answer options

  1. 0{,}2\,\text{кг}
  2. 0{,}05\,\text{кг}
  3. 0{,}3\,\text{кг}
  4. 0{,}1\,\text{кг}
Show the walkthroughHide the walkthrough
Given
P
1130\,\text{Вт}
t
200\,\text{с}
L
2260000\,\text{Дж/кг}
Find

m

  1. Q=PtQ=1130 \cdot 200Q=226000\,\text{Дж}

    We count the energy transferred to boiling water.

  2. m=\frac{Q}{L}m=\frac{226000}{2260000}m=0{,}1\,\text{кг}

    All the heat is spent on steam formation.

Answer0{,}1\,\text{кг}

Position in this set: 25 from 27

Problem #112

Boiling and vaporization

Boiling water transmit power 2260 W B current 300 s. Which one mass will evaporate at absence losses? Specific latent heat of vaporization 2260000 J/kg.

Answer options

  1. 0{,}3\,\text{кг}
  2. 0{,}6\,\text{кг}
  3. 0{,}15\,\text{кг}
  4. 0{,}9\,\text{кг}
Show the walkthroughHide the walkthrough
Given
P
2260\,\text{Вт}
t
300\,\text{с}
L
2260000\,\text{Дж/кг}
Find

m

  1. Q=PtQ=2260 \cdot 300Q=678000\,\text{Дж}

    We count the energy transferred to boiling water.

  2. m=\frac{Q}{L}m=\frac{678000}{2260000}m=0{,}3\,\text{кг}

    All the heat is spent on steam formation.

Answer0{,}3\,\text{кг}

Position in this set: 26 from 27

Problem #113

Boiling and vaporization

water mass 1 kg heated up from 20 up to 100 °C, then evaporated 0{,}1 kg. Find full warmth. c = 4200 J/(kg °C), L = 2260000 J/kg, losses no.

Answer options

  1. 562000\,\text{Дж}
  2. 1124000\,\text{Дж}
  3. 281000\,\text{Дж}
  4. 1686000\,\text{Дж}
Show the walkthroughHide the walkthrough
Given
m
1\,\text{кг}
T
20\,\text{°C}
v
0{,}1\,\text{кг}
c
4200\,\text{Дж/(кг·°C)}
L
2260000\,\text{Дж/кг}
Find

Q

  1. Q_1=cm(100-T)Q_1=4200 \cdot 1 \cdot (100 - 20)Q_1=336000\,\text{Дж}

    All the water is heated to boiling.

  2. Q_2=LvQ_2=2260000 \cdot 0{,}1Q_2=226000\,\text{Дж}

    Only the specified mass is used for vaporization.

  3. Q=Q_1+Q_2Q=336000 + 226000Q=562000\,\text{Дж}

    Let's sum up the heat of the two stages.

Answer562000\,\text{Дж}

Position in this set: 27 from 27

Problem #114

Boiling and vaporization

water mass 2 kg heated up from 30 up to 100 °C, then evaporated 0{,}2 kg. Find full warmth. c = 4200 J/(kg °C), L = 2260000 J/kg, losses no.

Answer options

  1. 2080000\,\text{Дж}
  2. 1040000\,\text{Дж}
  3. 520000\,\text{Дж}
  4. 3120000\,\text{Дж}
Show the walkthroughHide the walkthrough
Given
m
2\,\text{кг}
T
30\,\text{°C}
v
0{,}2\,\text{кг}
c
4200\,\text{Дж/(кг·°C)}
L
2260000\,\text{Дж/кг}
Find

Q

  1. Q_1=cm(100-T)Q_1=4200 \cdot 2 \cdot (100 - 30)Q_1=588000\,\text{Дж}

    All the water is heated to boiling.

  2. Q_2=LvQ_2=2260000 \cdot 0{,}2Q_2=452000\,\text{Дж}

    Only the specified mass is used for vaporization.

  3. Q=Q_1+Q_2Q=588000 + 452000Q=1040000\,\text{Дж}

    Let's sum up the heat of the two stages.

Answer1040000\,\text{Дж}

The theory behind these problems

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

Melting, solidification and graphsMelting temperature and energy of a phase change.Evaporation, condensation and humiditySaturated vapour and measuring relative humidity.Boiling and vaporizationBoiling temperature, pressure and specific latent heat of vaporization.

Formulas by topic

Look up symbols and units of measurement in the reference.

Densityρ · кг/м³