## Related questions with answers

Heat pumps an be used as a practical way of heating buildings. The ground itself can be used as the cold source because at a depth of a few metres the temperature is independent of the air temperature; in temperate latitudes the ground temperature is around $13^{\circ} \mathrm{C}$ at a depth of 10 m. On a cold day it is found that to keep a certain room at $18^{\circ} \mathrm{C}$ a heater rated at 5 kW is required. Assuming that an ideal heat pump is used. and that all heat transfers are mersible, alculate the power needed to maintain the room temperature. Recall that $1 \mathrm{W}=1 \mathrm{J\ s}^{-1}.$

Solutions

Verified$\begin{aligned} \text{Given,}\\ \text{Temperature of cold source,}T_{c}&=13\degree C\\ \text{Temperature of room(hot sink),}T_{h}&=18\degree C\\ \text{Power of heater,P}&=5\text{ kW}\\ P&=\left(5\text{ kW}\right)\left(\dfrac{1000\text{ W}}{1\text{ kW}}\right)\left(\dfrac{1\text{ Js}^{-1}}{1\text{ W}}\right)\\ &=5\cdot 10^{3}\text{ Js}^{-1} \end{aligned}$

The magnitude of work need to make transfer heat of magnitude q+w from a cold source to hot sink can be given by,

$\begin{aligned} \vert w\vert&=\vert q\vert\left(1-\dfrac{T_{h}}{T_{c}}\right)\\ \text{Where,}\\ \vert q\vert&=\text{magnitude of heat extracted from cold source}\\ \text{Therefore,}\\ \vert w\vert&=\left(5\cdot 10^{3}\right)\left(1-\dfrac{18\degree C}{13\degree C}\right)\\ &=\left(5\cdot 10^{3}\right)\left(-0.3846\right)\\ &=\vert-1.923\cdot 10^{3}\text{ J}\vert\\ &=\boxed{1.923\cdot 10^{3}\text{ J}}\\ \end{aligned}$

The work done per second to maintain the room temperature

$\begin{aligned} &=\boxed{1.923\cdot 10^{3}\text{ J}}\\ \end{aligned}$

Therefore, the corresponding power needed to maintain the room temperature

$\begin{aligned} &=\boxed{1.923\cdot 10^{3}\text{ W}}\\ \end{aligned}$

Temperature of source, $T_c=13^\circ\mathrm {C}$ Temperature of room, $T_h=18^\circ \mathrm{C}$ Heater power, $P=5\ \mathrm{kW}$

$P=5\ \mathrm{kW}\cdot \frac{1000\ \mathrm{W}}{1\ \mathrm{kW}}\cdot \frac{1\ \mathrm{J/s}}{1\ \mathrm{W}}=5\cdot 10^{-3}\ \mathrm{J/s}=\lvert q\rvert$

$\lvert q\rvert= magnitude\space of\space exchanged\space heat$

Work needed to transfer heat (q+w) from source to room is:

$\begin{aligned} \lvert w\rvert &=\lvert q\rvert\cdot\bigg(1-\frac{T_h}{T_c}\bigg)\\\ &= 5\cdot 10^{3} \ \mathrm{J/s}\cdot\bigg (1-\frac{18^{\circ\mathrm{ C}}}{13^{\circ \mathrm{C}}}\bigg)\\\\ &=(5\cdot 10^{3}\cdot (-0.3846)\\\\ &=\lvert -1.923\cdot 10^{3}\ \mathrm{J}\rvert\\\\ &=1.923\cdot 10^{3} \ \mathrm{J}= The\space work\space done /s\space for\space maintenance\\\ \end{aligned}$

The power needed for maintaining room temperature:

$=1.923\cdot 10^{3}\ \mathrm{ W}$

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