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GRADE 10 PRACTICE

Physics Problems: Thermodynamics, Grade 10

Internal energy, heat, work and cycles.

24 problems across 4 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.

24 problems
Position in this set: 1 from 24

Problem #3

Thermodynamics: internal energy and heat transfer

2\,\text{кг} ice taken from 0\,^{\circ}\text{C}, you need to melt it and heat the resulting water to 15\,^{\circ}\text{C}. How much heat will be required? Assume the specific heat of fusion of ice is 330 kJ/kg and the specific heat capacity of water is 4200 J/(kg K).

Answer options

  1. 7{,}86 \cdot 10^{6}\,\text{Дж}
  2. 7{,}86 \cdot 10^{5}\,\text{Дж}
  3. 3{,}93 \cdot 10^{5}\,\text{Дж}
  4. 1{,}572 \cdot 10^{6}\,\text{Дж}
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Given
m
2\,\text{кг}
t_1
0\,^{\circ}\text{C}
t_2
15\,^{\circ}\text{C}
\lambda
330\,000\,\tfrac{\text{Дж}}{\text{кг}}
c
4200\,\tfrac{\text{Дж}}{\text{кг}\cdot{}^{\circ}\text{C}}
Find

Q

  1. Q = Q_1 + Q_2 = m\lambda + mc\left(t_2 - t_1\right)

    First the ice melts, then the resulting water is heated.

  2. Q_1 = m\lambdaQ_1 = 2\,\text{кг}\cdot 330\,000\,\tfrac{\text{Дж}}{\text{кг}}Q_1 = 660\,000\,\text{Дж}
  3. Q_2 = mc\left(t_2 - t_1\right)Q_2 = 2\,\text{кг}\cdot 4200\,\tfrac{\text{Дж}}{\text{кг}\cdot{}^{\circ}\text{C}}\cdot 15\,^{\circ}\text{C}Q_2 = 126\,000\,\text{Дж}
  4. Q = Q_1 + Q_2Q = 660\,000\,\text{Дж} + 126\,000\,\text{Дж}Q = 786\,000\,\text{Дж}

AnswerQ = 786\,000\,\text{Дж}

Position in this set: 2 from 24

Problem #82

Thermodynamics: internal energy and heat transfer

How much heat is required to 2\,\text{кг} ice at -40\,^{\circ}\text{C} turn into water vapor 100\,^{\circ}\text{C}? Take the heat capacities of ice and water as 2100 and 4200 J/(kg K), the specific heat of melting of ice as 330 kJ/kg and the heat of vaporization of water as 2.3 MJ/kg.

Answer options

  1. 6{,}268 \cdot 10^{6}\,\text{Дж}
  2. 3{,}134 \cdot 10^{6}\,\text{Дж}
  3. 1{,}2536 \cdot 10^{7}\,\text{Дж}
  4. 6{,}268 \cdot 10^{7}\,\text{Дж}
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Given
m
2\,\text{кг}
t_1
-40\,^{\circ}\text{C}
t_2
0\,^{\circ}\text{C}
t_3
100\,^{\circ}\text{C}
c_1
2100\,\tfrac{\text{Дж}}{\text{кг}\cdot{}^{\circ}\text{C}}
\lambda
330\,000\,\tfrac{\text{Дж}}{\text{кг}}
c_2
4200\,\tfrac{\text{Дж}}{\text{кг}\cdot{}^{\circ}\text{C}}
r
2\,300\,000\,\tfrac{\text{Дж}}{\text{кг}}
Find

Q

  1. Q = mc_1\left(t_2 - t_1\right) + m\lambda + mc_2\left(t_3 - t_2\right) + mr

    Four stages: heating ice, melting, heating water, evaporation.

  2. Q = 2\cdot 2100\cdot 40 + 2\cdot 330\,000 + 2\cdot 4200\cdot 100 + 2\cdot 2\,300\,000Q = 168\,000 + 660\,000 + 840\,000 + 4\,600\,000Q = 6\,268\,000\,\text{Дж}

AnswerQ = 6\,268\,000\,\text{Дж}

Position in this set: 3 from 24

Problem #83

Thermodynamics: internal energy and heat transfer

Define internal energy 20\,\text{г} helium at 7\,^{\circ}\text{C}.

Answer options

  1. 1{,}7451 \cdot 10^{5}\,\text{Дж}
  2. 17451\,\text{Дж}
  3. 8725{,}5\,\text{Дж}
  4. 34902\,\text{Дж}
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Given
m
20\,\text{г}
t
7\,^{\circ}\text{C}T = 280\,\text{К}
M
4\,\tfrac{\text{г}}{\text{моль}}
R
8{,}31\,\tfrac{\text{Дж}}{\text{К}\cdot\text{моль}}
Find

U

  1. U = \dfrac{3}{2}\cdot\dfrac{m}{M}RT

    Helium is monatomic, so there are three degrees of freedom per molecule.

  2. U = \dfrac{3mRT}{2M}U = \dfrac{3\cdot 20\cdot 8{,}31\cdot 280}{2\cdot 4}U = 17\,451\,\text{Дж}

AnswerU = 17\,451\,\text{Дж}

Position in this set: 4 from 24

Problem #84

Thermodynamics: internal energy and heat transfer

How much alcohol do you need to burn to melt? 3\,\text{кг} lead taken from 127\,^{\circ}\text{C}? The melting point of lead is 327 °C, its specific heat capacity is 130 J/(kg K), the specific heat of fusion is 25 kJ/kg. The specific heat of combustion of alcohol is 29 MJ/kg. All the heat of combustion is transferred to lead.

Answer options

  1. 10{,}6\,\text{г}
  2. 53\,\text{г}
  3. 5{,}3\,\text{г}
  4. 2{,}65\,\text{г}
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Given
m
3\,\text{кг}
t_1
127\,^{\circ}\text{C}
t_2
327\,^{\circ}\text{C}
c
130\,\tfrac{\text{Дж}}{\text{кг}\cdot{}^{\circ}\text{C}}
\lambda
25\,000\,\tfrac{\text{Дж}}{\text{кг}}
q
29\cdot 10^{6}\,\tfrac{\text{Дж}}{\text{кг}}
Find

m_2

  1. Q = mc\left(t_2 - t_1\right) + m\lambda3\cdot 130\cdot 200 + 3\cdot 25\,000 = 78\,000 + 75\,000Q = 153\,000\,\text{Дж}

    Lead is first heated to its melting point and then melted.

  2. Q = m_2 q\ \Rightarrow\ m_2 = \dfrac{Q}{q}m_2 = \dfrac{153\,000}{29\cdot 10^{6}}m_2 = 5{,}3\cdot 10^{-3}\,\text{кг} = 5{,}3\,\text{г}

Answerm_2 = 5{,}3\,\text{г}

Position in this set: 5 from 24

Problem #85

Thermodynamics: heat engines

Hot reservoir temperature 327\,^{\circ}\text{C}, refrigerator -33\,^{\circ}\text{C}. What is the maximum efficiency? How much heat does a reversible Carnot engine receive from the hot reservoir if the heat rejected to the cold reservoir is 300\,\text{Дж}?

Answer options

  1. \left(0{,}3\,\text{доля};\ 375\,\text{Дж}\right)
  2. \left(1{,}2\,\text{доля};\ 1500\,\text{Дж}\right)
  3. \left(6\,\text{доля};\ 7500\,\text{Дж}\right)
  4. \left(0{,}6\,\text{доля};\ 750\,\text{Дж}\right)
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Given
t_1
327\,^{\circ}\text{C}T_1 = 600\,\text{К}
t_2
-33\,^{\circ}\text{C}T_2 = 240\,\text{К}
Q_2
300\,\text{Дж}
Find

\etaQ_1

  1. \eta = \dfrac{T_1 - T_2}{T_1}\eta = \dfrac{600 - 240}{600}\eta = 0{,}6

    The maximum efficiency is given by the Carnot cycle.

  2. \eta = \dfrac{Q_1 - Q_2}{Q_1}\ \Rightarrow\ Q_2 = Q_1\left(1 - \eta\right)
  3. Q_1 = \dfrac{Q_2}{1 - \eta}Q_1 = \dfrac{300}{1 - 0{,}6}Q_1 = 750\,\text{Дж}

Answer\eta = 0{,}6Q_1 = 750\,\text{Дж}

Position in this set: 6 from 24

Problem #86

Thermodynamics: internal energy and heat transfer

A monatomic gas is heated isobarically at pressure 2\cdot10^{5}\,\text{Па}. Volume increased by 15\,\text{л}. Determine the work done, the change in internal energy, and the heat gained.

Answer options

  1. \left(3000\,\text{Дж};\ 4500\,\text{Дж};\ 7500\,\text{Дж}\right)
  2. \left(1500\,\text{Дж};\ 2250\,\text{Дж};\ 3750\,\text{Дж}\right)
  3. \left(6000\,\text{Дж};\ 9000\,\text{Дж};\ 15000\,\text{Дж}\right)
  4. \left(30000\,\text{Дж};\ 45000\,\text{Дж};\ 75000\,\text{Дж}\right)
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Given
p
2\cdot 10^{5}\,\text{Па}
\Delta V
15\,\text{л}15\cdot 10^{-3}\,\text{м}^{3}
Find

A\Delta UQ

  1. A = p\,\Delta VA = 2\cdot 10^{5}\cdot 15\cdot 10^{-3}A = 3000\,\text{Дж}
  2. \Delta U = \dfrac{3}{2}\nu R\,\Delta T = \dfrac{3}{2}p\,\Delta V = \dfrac{3}{2}A\Delta U = \dfrac{3}{2}\cdot 3000\Delta U = 4500\,\text{Дж}
  3. Q = \Delta U + A = \dfrac{5}{2}AQ = \dfrac{5}{2}\cdot 3000Q = 7500\,\text{Дж}

    In an isobaric process, work accounts for two-fifths of the heat input.

AnswerA = 3000\,\text{Дж}\Delta U = 4500\,\text{Дж}Q = 7500\,\text{Дж}

Position in this set: 7 from 24

Problem #87

Thermodynamics: internal energy and heat transfer

The bath is filled 80\,\text{л} water at 15\,^{\circ}\text{C}. How much water at 80\,^{\circ}\text{C} need to be added to get 40\,^{\circ}\text{C}?

Answer options

  1. 500\,\text{л}
  2. 50\,\text{л}
  3. 25\,\text{л}
  4. 100\,\text{л}
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Given
m_1
80\,\text{кг}
t_1
15\,^{\circ}\text{C}
t_2
80\,^{\circ}\text{C}
\theta
40\,^{\circ}\text{C}
Find

V_2

  1. m_1 c\left(\theta - t_1\right) = m_2 c\left(t_2 - \theta\right)

    Heat balance equation; the specific heat capacity decreases.

  2. m_2 = \dfrac{m_1\left(\theta - t_1\right)}{t_2 - \theta}m_2 = \dfrac{80\cdot 25}{40}V_2 = 50\,\text{кг} \Rightarrow V_2 = 50\,\text{л}

AnswerV_2 = 50\,\text{л}

Position in this set: 8 from 24

Problem #88

Thermodynamics: internal energy and heat transfer

Into the water at 50\,^{\circ}\text{C} threw a piece of ice 20\,\text{г} at 0\,^{\circ}\text{C}. The temperature dropped to 20\,^{\circ}\text{C}. What is the initial mass of water? The specific heat of melting of ice is 330 kJ/kg, the heat capacity of water is 4200 J/(kg K). Neglect heat exchange with the environment.

Answer options

  1. 131{,}4\,\text{г}
  2. 657\,\text{г}
  3. 65{,}7\,\text{г}
  4. 32{,}85\,\text{г}
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Given
t_1
50\,^{\circ}\text{C}
m_2
20\,\text{г}0{,}02\,\text{кг}
t_2
0\,^{\circ}\text{C}
\theta
20\,^{\circ}\text{C}
c
4200\,\tfrac{\text{Дж}}{\text{кг}\cdot{}^{\circ}\text{C}}
\lambda
330\,000\,\tfrac{\text{Дж}}{\text{кг}}
Find

m_1

  1. m_1 c\left(t_1 - \theta\right) = m_2\lambda + m_2 c\left(\theta - t_2\right)

    The ice first melts, then the resulting water is heated to a common temperature.

  2. m_1 = \dfrac{m_2\lambda + m_2 c\left(\theta - t_2\right)}{c\left(t_1 - \theta\right)}m_1 = \dfrac{0{,}02\cdot 330\,000 + 0{,}02\cdot 4200\cdot 20}{4200\cdot 30}m_1 = 0{,}0657\,\text{кг} = 65{,}7\,\text{г}

Answerm_1 = 65{,}7\,\text{г}

Position in this set: 9 from 24

Problem #89

Thermodynamics: internal energy and heat transfer

In a glass of mass 100\,\text{г} poured 200\,\text{г} water at 30\,^{\circ}\text{C}. How much ice at -15\,^{\circ}\text{C} must be added to cool the system to 10\,^{\circ}\text{C}? Glass glass. Specific heat capacities of glass, water and ice: 830, 4200 and 2100 J/(kg K). The specific heat of melting of ice is 330 kJ/kg. Neglect heat exchange with the environment.

Answer options

  1. 23\,\text{г}
  2. 92\,\text{г}
  3. 460\,\text{г}
  4. 46\,\text{г}
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Given
m_1
0{,}1\,\text{кг}
m_2
0{,}2\,\text{кг}
t_1
30\,^{\circ}\text{C}
t_3
-15\,^{\circ}\text{C}
t_2
0\,^{\circ}\text{C}
\theta
10\,^{\circ}\text{C}
c_1
830\,\tfrac{\text{Дж}}{\text{кг}\cdot{}^{\circ}\text{C}}
c_2
4200\,\tfrac{\text{Дж}}{\text{кг}\cdot{}^{\circ}\text{C}}
c_3
2100\,\tfrac{\text{Дж}}{\text{кг}\cdot{}^{\circ}\text{C}}
\lambda
330\,000\,\tfrac{\text{Дж}}{\text{кг}}
Find

m_3

  1. m_1 c_1\left(t_1 - \theta\right) + m_2 c_2\left(t_1 - \theta\right) = m_3\left[c_3\left(t_2 - t_3\right) + \lambda + c_2\left(\theta - t_2\right)\right]

    On the left the glass and water give off heat, on the right the ice heats up, melts and heats up like water.

  2. m_3 = \dfrac{m_1 c_1\left(t_1 - \theta\right) + m_2 c_2\left(t_1 - \theta\right)}{c_3\left(t_2 - t_3\right) + \lambda + c_2\left(\theta - t_2\right)}m_3 = \dfrac{0{,}1\cdot 830\cdot 20 + 0{,}2\cdot 4200\cdot 20}{2100\cdot 15 + 330\,000 + 4200\cdot 10}m_3 = 4{,}6\cdot 10^{-2}\,\text{кг} = 46\,\text{г}

Answerm_3 = 46\,\text{г}

Position in this set: 10 from 24

Problem #90

Thermodynamics: internal energy and heat transfer

Mass into the water 2\,\text{кг} at 20\,^{\circ}\text{C} let me in 100\,\text{г} water vapor at 100\,^{\circ}\text{C}. What temperature will it be? The specific heat of vaporization of water is 2.3 MJ/kg, the heat capacity of water is 4200 J/(kg K). Neglect heat exchange with the environment.

Answer options

  1. 50\,^{\circ}\text{C}
  2. 25\,^{\circ}\text{C}
  3. 100\,^{\circ}\text{C}
  4. 500\,^{\circ}\text{C}
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Given
m_1
2\,\text{кг}
t_1
20\,^{\circ}\text{C}
m_2
100\,\text{г}0{,}1\,\text{кг}
t_2
100\,^{\circ}\text{C}
c
4200\,\tfrac{\text{Дж}}{\text{кг}\cdot{}^{\circ}\text{C}}
r
2\,300\,000\,\tfrac{\text{Дж}}{\text{кг}}
Find

\theta

  1. m_1 c\left(\theta - t_1\right) = m_2 r + m_2 c\left(t_2 - \theta\right)

    The steam first condenses, then the resulting water cools.

  2. \theta = \dfrac{m_2 r + m_2 c t_2 + m_1 c t_1}{m_1 c + m_2 c}\theta = \dfrac{0{,}1\cdot 2\,300\,000 + 0{,}1\cdot 4200\cdot 100 + 2\cdot 4200\cdot 20}{2\cdot 4200 + 0{,}1\cdot 4200}\theta = 50\,^{\circ}\text{C}

Answer\theta = 50\,^{\circ}\text{C}

Position in this set: 11 from 24

Problem #91

Thermodynamics: internal energy and heat transfer

Determine work during isobaric expansion 56\,\text{г} nitrogen, if the initial temperature 0\,^{\circ}\text{C}, and the volume increased by 3 times.

Answer options

  1. 90745\,\text{Дж}
  2. 9074{,}5\,\text{Дж}
  3. 4537{,}25\,\text{Дж}
  4. 18149\,\text{Дж}
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Given
m
56\,\text{г}
t_1
0\,^{\circ}\text{C}T_1 = 273\,\text{К}
V_2
3V_1
M
28\,\tfrac{\text{г}}{\text{моль}}
R
8{,}31\,\tfrac{\text{Дж}}{\text{К}\cdot\text{моль}}
Find

A

  1. \dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}\ \Rightarrow\ T_2 = \dfrac{V_2 T_1}{V_1} = 3T_1T_2 = 3\cdot 273T_2 = 819\,\text{К}
  2. A = p\,\Delta V = \dfrac{m}{M}R\left(T_2 - T_1\right)A = \dfrac{56\cdot 8{,}31\cdot 546}{28}A = 9074{,}5\,\text{Дж}

AnswerA = 9074{,}5\,\text{Дж}

Position in this set: 12 from 24

Problem #93

Thermodynamics: internal energy and heat transfer

In a steel mug weighing 500\,\text{г} there is water at 15\,^{\circ}\text{C}. A copper cylinder with a mass of 200\,\text{г} at 150\,^{\circ}\text{C}. After heat exchange the temperature became 20\,^{\circ}\text{C}. What is the mass of water? Take the specific heat capacities of steel, water and copper equal to 500, 4200 and 400 J/(kg K). Neglect heat exchange with the environment.

Answer options

  1. 0{,}22\,\text{кг}
  2. 0{,}88\,\text{кг}
  3. 4{,}4\,\text{кг}
  4. 0{,}44\,\text{кг}
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Given
m_1
500\,\text{г}0{,}5\,\text{кг}
t_1 = t_2
15\,^{\circ}\text{C}
m_3
200\,\text{г}0{,}2\,\text{кг}
t_3
150\,^{\circ}\text{C}
\theta
20\,^{\circ}\text{C}
c_1
500\,\tfrac{\text{Дж}}{\text{кг}\cdot{}^{\circ}\text{C}}
c_2
4200\,\tfrac{\text{Дж}}{\text{кг}\cdot{}^{\circ}\text{C}}
c_3
400\,\tfrac{\text{Дж}}{\text{кг}\cdot{}^{\circ}\text{C}}
Find

m_2

  1. m_1 c_1\left(\theta - t_1\right) + m_2 c_2\left(\theta - t_2\right) = m_3 c_3\left(t_3 - \theta\right)

    The mug and water heat up, the copper cylinder cools down.

  2. m_2 = \dfrac{m_3 c_3\left(t_3 - \theta\right) - m_1 c_1\left(\theta - t_1\right)}{c_2\left(\theta - t_2\right)}m_2 = \dfrac{0{,}2\cdot 400\cdot 130 - 0{,}5\cdot 500\cdot 5}{4200\cdot 5}m_2 = 0{,}44\,\text{кг}

Answerm_2 = 0{,}44\,\text{кг}

Position in this set: 13 from 24

Problem #195

Internal energy and the first law

Gas expands at constant pressure 100000 Pa, increasing the volume by 0{,}002 m³. Find the work done by the gas.

Answer options

  1. 200\,\text{Дж}
  2. 400\,\text{Дж}
  3. 100\,\text{Дж}
  4. 600\,\text{Дж}
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Given
p
100000\,\text{Па}
V
0{,}002\,\text{м}^{3}
Find

A

  1. A=p\Delta VA=100000 \cdot 0{,}002A=200\,\text{Дж}

    During expansion, the work done by the gas is positive.

Answer200\,\text{Дж}

Position in this set: 14 from 24

Problem #196

Internal energy and the first law

Gas expands at constant pressure 200000 Pa, increasing the volume by 0{,}003 m³. Find the work done by the gas.

Answer options

  1. 1200\,\text{Дж}
  2. 600\,\text{Дж}
  3. 300\,\text{Дж}
  4. 1800\,\text{Дж}
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Given
p
200000\,\text{Па}
V
0{,}003\,\text{м}^{3}
Find

A

  1. A=p\Delta VA=200000 \cdot 0{,}003A=600\,\text{Дж}

    During expansion, the work done by the gas is positive.

Answer600\,\text{Дж}

Position in this set: 15 from 24

Problem #197

Internal energy and the first law

Gas received 1000 J of heat and expanded from 0{,}002 up to 0{,}005 m³ at constant pressure 100000 Pa. Find the change in internal energy.

Answer options

  1. 1400\,\text{Дж}
  2. 700\,\text{Дж}
  3. 350\,\text{Дж}
  4. 2100\,\text{Дж}
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Given
Q
1000\,\text{Дж}
V_1
0{,}002\,\text{м}^{3}
V_2
0{,}005\,\text{м}^{3}
p
100000\,\text{Па}
Find

\Delta U

  1. A=p(V_2-V_1)A=100000 \cdot (0{,}005 - 0{,}002)A=300\,\text{Дж}

    We find the work done by the gas itself.

  2. \Delta U=Q-A\Delta U=1000 - 300\Delta U=700\,\text{Дж}

    The remaining energy increases internal energy.

Answer700\,\text{Дж}

Position in this set: 16 from 24

Problem #198

Internal energy and the first law

Gas received 1800 J of heat and expanded from 0{,}003 up to 0{,}007 m³ at constant pressure 200000 Pa. Find the change in internal energy.

Answer options

  1. 2000\,\text{Дж}
  2. 500\,\text{Дж}
  3. 1000\,\text{Дж}
  4. 3000\,\text{Дж}
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Given
Q
1800\,\text{Дж}
V_1
0{,}003\,\text{м}^{3}
V_2
0{,}007\,\text{м}^{3}
p
200000\,\text{Па}
Find

\Delta U

  1. A=p(V_2-V_1)A=200000 \cdot (0{,}007 - 0{,}003)A=800\,\text{Дж}

    We find the work done by the gas itself.

  2. \Delta U=Q-A\Delta U=1800 - 800\Delta U=1000\,\text{Дж}

    The remaining energy increases internal energy.

Answer1000\,\text{Дж}

Position in this set: 17 from 24

Problem #199

Internal energy and the first law

Got gas first 500 J at constant volume. Then I got more 900 J and expanded by 0{,}002 m³ at pressure 100000 Pa. Find the total change in internal energy.

Answer options

  1. 2400\,\text{Дж}
  2. 600\,\text{Дж}
  3. 1200\,\text{Дж}
  4. 3600\,\text{Дж}
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Given
Q_1
500\,\text{Дж}
Q_2
900\,\text{Дж}
V
0{,}002\,\text{м}^{3}
p
100000\,\text{Па}
Find

\Delta U

  1. A=p\Delta VA=100000 \cdot 0{,}002A=200\,\text{Дж}

    With a constant amount of work there is no; it arises only at the second stage.

  2. Q=Q_1+Q_2Q=500 + 900Q=1400\,\text{Дж}

    Add up the resulting heat.

  3. \Delta U=Q-A\Delta U=1400 - 200\Delta U=1200\,\text{Дж}

    We apply the first law to the entire process.

Answer1200\,\text{Дж}

Position in this set: 18 from 24

Problem #200

Internal energy and the first law

Got gas first 800 J at constant volume. Then I got more 1200 J and expanded by 0{,}003 m³ at pressure 200000 Pa. Find the total change in internal energy.

Answer options

  1. 2800\,\text{Дж}
  2. 700\,\text{Дж}
  3. 4200\,\text{Дж}
  4. 1400\,\text{Дж}
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Given
Q_1
800\,\text{Дж}
Q_2
1200\,\text{Дж}
V
0{,}003\,\text{м}^{3}
p
200000\,\text{Па}
Find

\Delta U

  1. A=p\Delta VA=200000 \cdot 0{,}003A=600\,\text{Дж}

    With a constant amount of work there is no; it arises only at the second stage.

  2. Q=Q_1+Q_2Q=800 + 1200Q=2000\,\text{Дж}

    Add up the resulting heat.

  3. \Delta U=Q-A\Delta U=2000 - 600\Delta U=1400\,\text{Дж}

    We apply the first law to the entire process.

Answer1400\,\text{Дж}

Position in this set: 19 from 24

Problem #201

Thermal processes and efficiency

Useful installation work 300 J at energy expended 500 J. Find the efficiency as a percentage.

Answer options

  1. 120\,\text{\%}
  2. 60\,\text{\%}
  3. 30\,\text{\%}
  4. 180\,\text{\%}
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Given
A
300\,\text{Дж}
Q
500\,\text{Дж}
Find

\eta

  1. \eta=\frac{A}{Q}\cdot100\eta=\frac{300}{500} \cdot 100\eta=60\,\text{\%}

    Useful energy does not exceed expended energy.

Answer60\,\text{\%}

Position in this set: 20 from 24

Problem #202

Thermal processes and efficiency

Useful installation work 450 J at energy expended 1000 J. Find the efficiency as a percentage.

Answer options

  1. 90\,\text{\%}
  2. 22{,}5\,\text{\%}
  3. 45\,\text{\%}
  4. 135\,\text{\%}
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Given
A
450\,\text{Дж}
Q
1000\,\text{Дж}
Find

\eta

  1. \eta=\frac{A}{Q}\cdot100\eta=\frac{450}{1000} \cdot 100\eta=45\,\text{\%}

    Useful energy does not exceed expended energy.

Answer45\,\text{\%}

Position in this set: 21 from 24

Problem #203

Thermal processes and efficiency

Installation received 2000 J of energy and has efficiency 60 %. How much energy was wasted?

Answer options

  1. 1600\,\text{Дж}
  2. 400\,\text{Дж}
  3. 800\,\text{Дж}
  4. 2400\,\text{Дж}
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Given
Q
2000\,\text{Дж}
e
60\,\text{\%}
Find

R

  1. A=Q\frac{\eta}{100}A=\frac{2000 \cdot 60}{100}A=1200\,\text{Дж}

    We convert the percentage of efficiency into a share.

  2. R=Q-AR=2000 - 1200R=800\,\text{Дж}

    The losses account for the remaining energy.

Answer800\,\text{Дж}

Position in this set: 22 from 24

Problem #204

Thermal processes and efficiency

Installation received 3000 J of energy and has efficiency 70 %. How much energy was wasted?

Answer options

  1. 1800\,\text{Дж}
  2. 450\,\text{Дж}
  3. 2700\,\text{Дж}
  4. 900\,\text{Дж}
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Given
Q
3000\,\text{Дж}
e
70\,\text{\%}
Find

R

  1. A=Q\frac{\eta}{100}A=\frac{3000 \cdot 70}{100}A=2100\,\text{Дж}

    We convert the percentage of efficiency into a share.

  2. R=Q-AR=3000 - 2100R=900\,\text{Дж}

    The losses account for the remaining energy.

Answer900\,\text{Дж}

Position in this set: 23 from 24

Problem #205

Thermal processes and efficiency

Installation receives 10000 J. The first stage passes on 80 % of energy, the second turns into useful work 50 % of energy received by it. Find the total energy loss.

Answer options

  1. 12000\,\text{Дж}
  2. 3000\,\text{Дж}
  3. 18000\,\text{Дж}
  4. 6000\,\text{Дж}
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Given
Q
10000\,\text{Дж}
e_1
80\,\text{\%}
e_2
50\,\text{\%}
Find

R

  1. E=Q\frac{\eta_1}{100}E=\frac{10000 \cdot 80}{100}E=8000\,\text{Дж}

    In the second stage, only the energy after the first is available.

  2. A=E\frac{\eta_2}{100}A=\frac{8000 \cdot 50}{100}A=4000\,\text{Дж}

    The efficiencies of the stages are multiplied, not added.

  3. R=Q-AR=10000 - 4000R=6000\,\text{Дж}

    Subtract the final useful work from the initial energy.

Answer6000\,\text{Дж}

Position in this set: 24 from 24

Problem #206

Thermal processes and efficiency

Installation receives 20000 J. The first stage passes on 75 % of energy, the second turns into useful work 60 % of energy received by it. Find the total energy loss.

Answer options

  1. 11000\,\text{Дж}
  2. 22000\,\text{Дж}
  3. 5500\,\text{Дж}
  4. 33000\,\text{Дж}
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Given
Q
20000\,\text{Дж}
e_1
75\,\text{\%}
e_2
60\,\text{\%}
Find

R

  1. E=Q\frac{\eta_1}{100}E=\frac{20000 \cdot 75}{100}E=15000\,\text{Дж}

    In the second stage, only the energy after the first is available.

  2. A=E\frac{\eta_2}{100}A=\frac{15000 \cdot 60}{100}A=9000\,\text{Дж}

    The efficiencies of the stages are multiplied, not added.

  3. R=Q-AR=20000 - 9000R=11000\,\text{Дж}

    Subtract the final useful work from the initial energy.

Answer11000\,\text{Дж}

The theory behind these problems

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

Internal energy and the first lawInternal energy of a monatomic gas, work, and the first law of thermodynamics.Thermal processes and efficiencyHeating, phase changes, adiabatic processes, and heat engines.

Formulas by topic

Look up symbols and units of measurement in the reference.

HeatQ · ДжEfficiencyη · %WorkA · ДжTemperatureT, t · К, °CSpecific heat capacityc · Дж/(кг·К)EnergyE, W · Дж