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Question: The temperature of a room heated by a heater is \(20^\circ C\) when outside temperature is \( - 20^\...

The temperature of a room heated by a heater is 20∘C20^\circ C when outside temperature is −20∘C - 20^\circ C and it is 10∘C10^\circ C when the outside temperature is −40∘C - 40^\circ C. The temperature of the heater is:
A) 80∘C80^\circ C
B) 100∘C100^\circ C
C) 40∘C40^\circ C
D) 60∘C60^\circ C

Explanation

Solution

When heater gives heat to room then temperature of room increases and outside temperature of room is below then room so from room heat start to transfer to the outside of room from inside of room. In the study state the rate of heat transfer increases from heater to room so the rate of heat transfer also increases between room and outside.

Complete step by step answer:
To solve this question we use at the study state the rate of heat flow between heater and room is equal to the rate of heat flow between room to outside.
I.e. Rate of heat flow between heater and room ∝\propto rate of heat flow between room and outside of room.
⇒(dQdt)heater→room∝(dQdt)room→outside\Rightarrow {\left( {\dfrac{{dQ}}{{dt}}} \right)_{heater \to room}} \propto {\left( {\dfrac{{dQ}}{{dt}}} \right)_{room \to outside}} ......................... (1)
Let us assume at the study state the temperature of heater is Th{T_h} and temperature of room is Tr{T_r} and outside temperature To{T_o}

We know the rate of heat flow is proportional to the temperature difference so we can write for the heater and room.
⇒(dQdt)heater→room∝(Th−Tr)\Rightarrow {\left( {\dfrac{{dQ}}{{dt}}} \right)_{heater \to room}} \propto \left( {{T_h} - {T_r}} \right)
And rate of flow of heat between room and outside
⇒(dQdt)room→outside∝(Tr−To)\Rightarrow {\left( {\dfrac{{dQ}}{{dt}}} \right)_{room \to outside}} \propto \left( {{T_r} - {T_o}} \right)
From equation (1)
⇒(Th−Tr)∝(Tr−To)\Rightarrow \left( {{T_h} - {T_r}} \right) \propto \left( {{T_r} - {T_o}} \right)
⇒(Th−Tr)=k(Tr−To)\Rightarrow \left( {{T_h} - {T_r}} \right) = k\left( {{T_r} - {T_o}} \right)................ (2)

There are given two condition in question when room temperature Tr=20∘C{T_r} = 20^\circ C and outside temperature is To=−20∘C{T_o} = - 20^\circ C
Put these value in equation (2)
(Th−20)=k(20−(−20))\left( {{T_h} - 20} \right) = k\left( {20 - \left( { - 20} \right)} \right) ............ (3)
When Tr=10∘C{T_r} = 10^\circ C and To=−40∘C{T_o} = - 40^\circ C then from (2)
⇒(Th−10)=k(10−(−40))\Rightarrow \left( {{T_h} - 10} \right) = k\left( {10 - \left( { - 40} \right)} \right).................. (4)
Divide equation (3) by (4)
⇒(Th−20)=k(20−(−20))(Th−10)=k(10−(−40))\Rightarrow \dfrac{{\left( {{T_h} - 20} \right) = k\left( {20 - \left( { - 20} \right)} \right)}}{{\left( {{T_h} - 10} \right) = k\left( {10 - \left( { - 40} \right)} \right)}}
Solving this
⇒(Th−20)(Th−10)=4050\Rightarrow \dfrac{{\left( {{T_h} - 20} \right)}}{{\left( {{T_h} - 10} \right)}} = \dfrac{{40}}{{50}}
⇒5Th−100=4Th−40\Rightarrow 5{T_h} - 100 = 4{T_h} - 40
Further solving
⇒Th=100−40\Rightarrow {T_h} = 100 - 40
∴Th=60∘C\therefore {T_h} = 60^\circ C

Hence option (D) is correct.

Note: The rate of heat flow can be defined as amount of heat flow per unit time which is depends on given factors as we can know the formula of rate of heat flow
dQdt=kAT2−T1L\dfrac{{dQ}}{{dt}} = kA\dfrac{{{T_2} - {T_1}}}{L}
Where k⇒k \Rightarrow coefficient of thermal conductivity
A⇒A \Rightarrow Area
L⇒L \Rightarrow Length of medium
From this formula we can understand the rate of heat flow proportional to the temperature difference.