Question

The area of a glass of a window of a room is \[10{m^2}\] and thickness is \[2mm\] . The outer and inner temperatures are \[{40^{\rm O}}\] C and \[{20^{\rm O}}\] C respectively. The thermal conductivity of glass in the S.I unit is 0.2. The heat flowing in the room per second will be:
(A) \[3 \times {10^4}J\]
(B) \[2 \times {10^4}J\]
(C) \[30J\]
(D) \[45J\]

Answer 1

Hint: We know that the heat flowing through the room per second is given by \[\dfrac{Q}{t} = \dfrac{{A.K\Delta t}}{{dx}}\] . Here, \[\dfrac{Q}{t}\] is representing the heat flowing in the room per second, A is the area of the glass window, K is the thermal conductivity of glass in S.I unit, \[\Delta t\] is the temperature change, dx is the thickness of the glass window. We will put all the values and find the value of the flow of heat in the room per second.

Complete step by step answer
It is given in the question that the area of the glass window of a room is \[10{m^2}\] and thickness is \[2mm\] , the outer and temperature is \[{40^{\rm O}}\] C, and the inner temperature is \[{20^{\rm O}}\] C. it is also given that the thermal conductivity of glass in S.I unit is 0.2. Then we have to find the heat flowing in the room per second.
We know that the heat flowing through the room per second is given by \[\dfrac{Q}{t} = \dfrac{{A.K\Delta t}}{{dx}}\] .
Here, \[\dfrac{Q}{t}\] is representing the heat flowing in the room per second, A is the area of the glass window, K is the thermal conductivity of glass in S.I unit, \[\Delta t\] is the temperature change, dx is the thickness of the glass window.
Here we will convert the thickness of the glass window into the meter. We know that \[1m = {10^3}mm\] we get-
\[2mm = 2 \times {10^{ - 3}}m\].
The temperature change \[\Delta t\] is given by outer temperature – inner temperature. We get-
\[\Delta t = {40^{\rm O}} - {20^{\rm O}}\]\[ = {20^{\rm O}}\]C.
On putting the values of A, K, dx, \[\Delta t\] in \[\dfrac{Q}{t}\] , we get-
\[\dfrac{Q}{t} = \dfrac{{10 \times 0.2 \times 20}}{{2 \times {{10}^{ - 3}}}}\].
Here the value of t in L.H.S is one second, we get-
\[Q = \dfrac{{10 \times 0.2 \times 20}}{{2 \times {{10}^{ - 3}}}}\]
\[Q = \dfrac{{10 \times 2 \times 2}}{{0.0002}}\]
\[Q = \dfrac{{40}}{{0.0002}}\]\[ = 20000J\]
\[Q = 2 \times {10^4}J\].
So, the flow of heat in the room per second will be \[Q = 2 \times {10^4}J\] .

Therefore, option B is correct.

Additional information
We have converted the value of dx in meters because the value thermal conductivity is present in the D.I unit. The S.I unit of length is a meter so, we have converted the value of dx i.e., the thickness of the glass window in the meter.
The thermal conductivity is affected by change in temperature . A sharp change is observed when the temperature approaches zero kelvin which is applicable for metals and for non metals when the temperature is decreased below a certain temperature called debye temperature the thermal conductivity decreases along with heat capacity.

Note
One can make a mistake in the calculation, they may take the value of dx as $2mm$ directly which will give us the different value of the rate of flow of heat. To avoid this, we have to change the value of dx into the meter.