Answer

Verified

87k+ views

**Hint:**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.

$ \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 ${T_h}$ and temperature of room is ${T_r}$ and outside temperature ${T_o}$

We know the rate of heat flow is proportional to the temperature difference so we can write for the heater and room.

$ \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

$ \Rightarrow {\left( {\dfrac{{dQ}}{{dt}}} \right)_{room \to outside}} \propto \left( {{T_r} - {T_o}} \right)$

From equation (1)

$ \Rightarrow \left( {{T_h} - {T_r}} \right) \propto \left( {{T_r} - {T_o}} \right)$

$ \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 ${T_r} = 20^\circ C$ and outside temperature is ${T_o} = - 20^\circ C$

Put these value in equation (2)

$\left( {{T_h} - 20} \right) = k\left( {20 - \left( { - 20} \right)} \right)$ ............ (3)

When ${T_r} = 10^\circ C$ and ${T_o} = - 40^\circ C$ then from (2)

$ \Rightarrow \left( {{T_h} - 10} \right) = k\left( {10 - \left( { - 40} \right)} \right)$.................. (4)

Divide equation (3) by (4)

$ \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

$ \Rightarrow \dfrac{{\left( {{T_h} - 20} \right)}}{{\left( {{T_h} - 10} \right)}} = \dfrac{{40}}{{50}}$

$ \Rightarrow 5{T_h} - 100 = 4{T_h} - 40$

Further solving

$ \Rightarrow {T_h} = 100 - 40$

$\therefore {T_h} = 60^\circ C$

**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

$\dfrac{{dQ}}{{dt}} = kA\dfrac{{{T_2} - {T_1}}}{L}$

Where $k \Rightarrow $ coefficient of thermal conductivity

$A \Rightarrow $ Area

$L \Rightarrow $ Length of medium

From this formula we can understand the rate of heat flow proportional to the temperature difference.

Recently Updated Pages

Name the scale on which the destructive energy of an class 11 physics JEE_Main

Write an article on the need and importance of sports class 10 english JEE_Main

Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main

Choose the one which best expresses the meaning of class 9 english JEE_Main

What does a hydrometer consist of A A cylindrical stem class 9 physics JEE_Main

A motorcyclist of mass m is to negotiate a curve of class 9 physics JEE_Main

Other Pages

A pilot in a plane wants to go 500km towards the north class 11 physics JEE_Main

A passenger in an aeroplane shall A Never see a rainbow class 12 physics JEE_Main

A circular hole of radius dfracR4 is made in a thin class 11 physics JEE_Main

The potential energy of a certain spring when stretched class 11 physics JEE_Main

The ratio of speed of sound in Hydrogen to that in class 11 physics JEE_MAIN

A roller of mass 300kg and of radius 50cm lying on class 12 physics JEE_Main