Question

A room at $20{}^\circ C$ is heated by a heater of resistance 20 ohm connected to 200 V mains. The temperature is uniform throughout the room and the heat is transmitted through a glass window of area $1\,{{m}^{2}}$ and thickness 0.2 cm. Calculate the temperature outside. Thermal conductivity of glass is $0.2\,cal/m{}^\circ C$ and the mechanical equivalent of heat is 4.2 j/cal.
A. $13.69{}^\circ C$
B. $15.24{}^\circ C$
C. $17.85{}^\circ C$
D. $19.96{}^\circ C$

Answer 1

Hint:There is a heater which produces heat. This heat is uniformly spread through the room with a window. We have to calculate the temperature outside the room. We can use an equation showing the relation between temperature, thermal conductivity with the rate of flow of heat to solve this problem.

Formula used:
Rate of heat flow is:
\[\dfrac{dQ}{dt}=kA\dfrac{\left( {{\theta }_{1}}-{{\theta }_{2}} \right)}{L}\]
Where Q is the heat flowing, k is the thermal conductivity of the material, A is the area and L be the length of the conductor and \[\left( {{\theta }_{1}}-{{\theta }_{2}} \right)\] be the difference in temperature.
And also, we use the equation of power for finding the heat produced by the heater in the room.
$P=\dfrac{{{V}^{2}}}{R}\dfrac{J}{s}=\dfrac{{{V}^{2}}}{R}\dfrac{cal}{s}$
Where P is the power and V is the voltage supplied and R is the resistance offered.

Complete step by step solution:
There is a room containing a heater which produces some heat. The voltage supply of the heater is 200V and it offers a resistance of 20 ohm. The heat produced is transmitted through a window of area $1{{m}^{2}}$ and thickness 0.2 cm.

Given, thermal conductivity of glass is $0.2cal/m{}^\circ C$ and mechanical equivalent of heat is 4.2 j/cal. Temperature of the room containing the heater is given. It can be written as:
${{\theta }_{1}}=20{}^\circ C$
We have to find the temperature outside the room it is taken as:
${{\theta }_{2}}=\theta $
Now applying equation of rate of flow of heat, we get
\[\dfrac{dQ}{dt}=kA\dfrac{\left( {{\theta }_{1}}-{{\theta }_{2}} \right)}{L} \\ \]
That is,
\[\dfrac{dQ}{dt}=1\times 0.2\times \dfrac{\left( 20-\theta \right)}{0.2\times {{10}^{-2}}}cal=100\left( 20-\theta \right)\]

Now heat produced per second by the heater in the room is:
\[P=\dfrac{{{200}^{2}}}{20\times 4.2}=476.2\dfrac{cal}{s}\]
But in this question, it is given that heat produced per second by heater is equal to the heat conducted by the window.
Therefore, we can write it as:
$100(20-\theta )=476.2$
Therefore, temperature outside the room is:
$\therefore \theta =15.24{}^\circ C$

Therefore, the answer is option B.

Note: In this question, unit conversion is important and also you should know how to convert joules to calories. Moreover, thermal conductivity refers to an object's capacity to conduct or transfer heat. Heat sinks are made of high thermal conductivity objects, while thermal insulators are made of low thermal conductivity materials.