A room at `20^@C` is heated by a heater of resistence 20 ohm connected to 200 VV mains. The temperature is uniform throughout the room and the heati s transmitted through a glass window of area `1m^2` and thickness 0.2 cm. Calculate the temperature outside. Thermal conductivity of glass is `0.2 cal//mC^@` s and mechanical equivalent of heat is `4.2 J//cal`.

Options

A. `13.69^(@)C`
B. `15.24^(@)C`
C. `17.85^(@)C`
D. `19.96^(@)C`

Answer

B

Explanation

If `theta` is the temperture of outside heat passing per second through the glass window
`(dQ)/(dt)=KA((theta_(1)-theta_(2)))/L`
`=(0.2xx1xx(20-theta)cal)/(0.2xx10^(-2))=100(20-theta)`
And heat produced pr second by the heater in the room
`P=(V^(2))/R`
`=(200xx200)/(20xx4.2)=476.2 (cal)/s`
Now as the temperature of the room is constant, the heat produced per second by heater must be equal to the heat conducted through the glass window.
`100(20 -theta)=476.2theta=15.24^(@)C`