Solveeit Logo
Login

Question

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

The temperature of a room heated by a radiator 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 radiator is
A) 50∘C50^\circ C
B) 70∘C70^\circ C
C) 90∘C90^\circ C
D) 60∘C60^\circ C

Explanation

Solution

When radiator gives heat to room then temperature of room increase 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 study state the rate of heat transfer as increase from radiator to room then rate of heat transfer also increase between room and outside

Step by step solution:
To solve this question we use at the study state the rate of heat flow between radiator and room is equal to the rate of heat flow between room to outside.
I.e. Rate of heat flow between radiator and room ∝\propto rate of heat flow between room and outside of room
⇒(dQdt)radiator→room∝(dQdt)room→outside\Rightarrow {\left( {\dfrac{{dQ}}{{dt}}} \right)_{radiator \to room}} \propto {\left( {\dfrac{{dQ}}{{dt}}} \right)_{room \to outside}} ......................... (1)
Let us assume at the study state the temperature of heater is Trad{T_{rad}} 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 heater and room
⇒(dQdt)radiator→room∝(Trad−Tr)\Rightarrow {\left( {\dfrac{{dQ}}{{dt}}} \right)_{radiator \to room}} \propto \left( {{T_{rad}} - {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)
⇒(Trad−Tr)∝(Tr−To)\Rightarrow \left( {{T_{rad}} - {T_r}} \right) \propto \left( {{T_r} - {T_o}} \right)
⇒(Trad−Tr)=k(Tr−To)\Rightarrow \left( {{T_{rad}} - {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)
(Trad−20)=k(20−(−20))\left( {{T_{rad}} - 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)
⇒(Trad−20)=k(20−(−20))(Trad−10)=k(10−(−40))\Rightarrow \dfrac{{\left( {{T_{rad}} - 20} \right) = k\left( {20 - \left( { - 20} \right)} \right)}}{{\left( {{T_{rad}} - 10} \right) = k\left( {10 - \left( { - 40} \right)} \right)}}
Solving this
⇒(Trad−20)(Trad−10)=4050\Rightarrow \dfrac{{\left( {{T_{rad}} - 20} \right)}}{{\left( {{T_{rad}} - 10} \right)}} = \dfrac{{40}}{{50}}
⇒5Trad−100=4Trad−40\Rightarrow 5{T_{rad}} - 100 = 4{T_{rad}} - 40
Further solving
⇒Trad=100−40\Rightarrow {T_{rad}} = 100 - 40
∴Trad=60∘C\therefore {T_{rad}} = 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.