Question

A body of length 1 m having a cross-sectional area \[0.75{m^2}\]has heat flow through it at the rate of \[6000\,J{s^{ - 1}}\] then find the temperature difference if \[K = 200J{m^{ - 1}}{K^{ - 1}}\].
A. \[{20^0}C\]
B. \[{40^0}C\]
C. \[{80^0}C\]
D. \[{100^0}C\]

Answer 1

Hint:In order to solve this problem we need to understand the rate of heat transfer. The rate of flow of heat is the amount of heat that is transferred per unit of time. Here, using the formula for heat flow we are going to find the solution for the temperature difference.

Formula Used:
To find the heat flow rate the formula is given as,
\[\dfrac{Q}{t} = KA\dfrac{{\Delta T}}{L}\]
Where,
A is cross-sectional area of a body
\[\Delta T\] is temperature difference between two ends of a body
L is length of a body
K is thermal conductivity of a body

Complete step by step solution:
Consider a body of length, \[l = 1m\]having a cross-sectional area of\[A = 0.75\,{m^2}\]. It has a heat flow through it at the rate of \[\dfrac{Q}{t} = 6000\,J{s^{ - 1}}\] then we need to find the temperature difference if \[K = 200\,J{m^{ - 1}}{K^{ - 1}}\].

The rate of flow of heat is,
\[\dfrac{Q}{t} = KA\dfrac{{\Delta T}}{L}\]
Rewrite the equation for \[\Delta T\] we obtain,
\[\Delta T = \left( {\dfrac{Q}{t}} \right)\dfrac{L}{{KA}}\]
Substitute the values of length, rate of heat flow, area, and thermal conductivity in the above equation, then we obtain as follows,
\[\Delta T = 6000 \times \dfrac{1}{{200 \times 0.75}}\]
\[\therefore \Delta T = {40^0}C\]
Therefore, the temperature difference of a body is found to be \[{40^0}C\].

Hence, option B is the correct answer.

Note:Here, in this question, we found the temperature difference using the equation for the heat flow as discussed above. Temperature difference will be obtained by subtracting two different isolated temperature readings, or by measuring the amount of rise of temperature or the fall of the temperature.