Answer to Question #141311 in Molecular Physics | Thermodynamics for Sridhar

Question

A heating element of mass 100g and having specific heat of 1J/(g°C) is exposed to surrounding air at 27°C .The element attains a steady state temperature of 127°C while absorbing 100W of electric power.If the power is switched off then approximate time taken by the element to cool down to 126°C will be
(Neglect radiation)
Ans 1s

Answer 1

According to the Newton's law of cooling, the rate of heat transfer out of the body in the steady state is:

$$P_{out} = \alpha (T_{steady} - T_{air})$$

where $\alpha$ is some constant, and $T_{steady} = 127\degree C$ and $T_{air} = 27\degree C$. In the steady state this rate is equal to the rate of heat transfer into the body, which is $P_{in} = 100W$. Thus, obtain:

$$P_{out} = P_{in}\\$$ $$\alpha (T_{steady} - T_{air}) = P_{in}\\ \alpha = \dfrac{P_{in}}{(T_{steady} - T_{air})} = \dfrac{100W}{100\degree C} = 1\dfrac{W}{\degree C}$$

After switching off the power, the amount of heat the heating element gave while cooling from $T_1 = 127\degree C$ to $T_2 = 126\degree C$ is:

$$Q_{cooling} = mc(T_1 - T_2)$$

where $m = 100g$ and $c = 1\dfrac{J}{g\cdot \degree C}$. Thus, obtain:

$$Q_{cooling} = 100g\cdot1\dfrac{J}{g\cdot \degree C}\cdot (127\degree C - 126\degree C) = 100J$$

On the other hand, the rate of heat loss, according to the Newton's law of cooling, is:

$$\dfrac{Q_{cooling}}{t} = \alpha(T_2 - T_{air})$$

where $t$ is the time of cooling and $T_2 = 127\degree C$ is taken as a body temperature during this process (since it haven't changed much). Thus, expressing $t$, obtain:

$$t = \dfrac{Q_{cooling}}{ \alpha(T_2 - T_{air})} = \dfrac{100J}{1\frac{W}{\degree C}\cdot (127\degree C - 27\degree C)} = 1s$$

Answer. 1s.