Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 15 - The Laws of Thermodynamics - General Problems - Page 441: 60

Answer

a) $5.05\times 10^4\;\rm J$ b) $32\;\rm min$

Work Step by Step

a) First of all, we need to find the removed heat from the water to cool it from 25$^\circ$C to -17$^\circ$C. $$Q_L=\overbrace{ m_{water}c_{water}\Delta T_1}^{{\color{red}\;\text{cooling water to 0 C}}}+\overbrace{ mL_f}^{{\color{red}\;\text{converting water into ice}}}+\overbrace{ m_{ice}c_{ice}\Delta T_2 }^{{\color{red}\;\text{Cooling ice to -17 C}}}$$ whereas $\Delta T_1$ is the temperature difference from the room temperature of 25$^\circ$C to the freezing point 0$^\circ$C, $\Delta T_2$ is the temperature difference from the freezing point of 0$^\circ$C to -17$^\circ$C, $L_f$ is the latent heat of fusion required to convert the whole amount of water into ice. Now we need to plug the given; $$Q_L=\left[0.65\cdot 4186\cdot 25\right]+[0.65\cdot 333\times 10^3]+\left[0.65\cdot 2100\cdot 17\right]$$ $$Q_L=3.077\times10^5\;\rm J\tag 1$$ Now we need to find the work by using the Carnot efficiency $$e=\dfrac{T_H-T_L}{T_H}=\dfrac{W}{Q_H}$$ whereas $Q_H=Q_L+W$; $$ \dfrac{T_H-T_L}{T_H}=\dfrac{W}{Q_L+W}$$ Solving for $W$; $$\left( \dfrac{T_H-T_L}{T_H}\right)\left(Q_L+W\right)= W$$ $$ Q_L\dfrac{T_H-T_L}{T_H}+W\dfrac{T_H-T_L}{T_H} = W$$ $$W-W\dfrac{T_H-T_L}{T_H}=Q_L\dfrac{T_H-T_L}{T_H}$$ $$W\left(1-\dfrac{T_H-T_L}{T_H}\right)=Q_L\dfrac{T_H-T_L}{T_H}$$ $$W\left(1-\left[1-\dfrac{ T_L}{T_H}\right]\right)=Q_L\dfrac{T_H-T_L}{T_H}$$ $$W \dfrac{ T_L}{T_H} =Q_L\dfrac{T_H-T_L}{T_H}$$ $$W =Q_L\dfrac{T_H-T_L}{T_H}\cdot \dfrac{T_H}{ T_L}$$ $$W =Q_L\dfrac{T_H-T_L}{ T_L}$$ Plugging $Q_L$ from (1) and plugging the known; (Remember to convert the temperatures into Kelvin). $$W =3.077\times10^5\cdot \dfrac{298-256}{ 256}$$ $$W=\color{red}{\bf 5.05\times 10^4}\;\rm J$$ b) We know that the power is given by $$P=\dfrac{W}{\Delta t}$$ Solving for $\Delta t$ to find the time interval the compressor takes to complete the process. $$\Delta t=\dfrac{W}{P}=\dfrac{5.05\times 10^4}{105}=481\;\rm s$$ and since the compressor works only 25$\%$ of the time, so the whole time it takes is 4 times if he worked without a stop. So that; $$\Delta t_{tot}=\dfrac{481}{0.25}=1924\;\rm s\approx \color{red}{\bf32}\;\rm min$$
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