Question

A metal ball of mass 1 kg is heated by means of a 20 W heater in a room at 20°C. The temperature of the ball becomes steady at 50°C. (a) Find the rate of loss of heat to the surrounding when the ball is at 50°C. (b) Assuming Newton's law of cooling, calculate the rate of loss of heat to the surrounding when the ball rises 30°C. (c) Assume that the temperature of the ball rises uniformly from 20°C to 30°C in 5 minutes. Find the total loss of heat to the surrounding during this period. (d) Calculate the specific heat capacity of the metal.

Answer 1

In steady state, the body has reached equilibrium. So, no more heat will be exchanged between the body and the surrounding.

This implies that at steady state,
Rate of loss of heat = Rate at which heat is supplied

Given:
Mass, m = 1 kg
Power of the heater = 20 W
Room temperature = 20°C

(a)At steady state,
Rate of loss of heat = Rate at which heat is supplied
And, rate of loss/gain of heat = Power

$\therefore \frac{\mathrm{dQ}}{\mathrm{d}t}=P=20\mathrm{W}$

(b) By Newton's law of cooling, rate of cooling is directly proportional to the difference in temperature.
So, when the body is in steady state, then its rate of cooling is given as

$$\begin{gathered} \frac{\mathrm{dQ}}{\mathrm{d}t}=K\left(T-T_0\right) \\ 20=K\left(50-20\right) \\ \Rightarrow K=\frac{2}{3} \end{gathered}$$
When the temperature of the body is 30°C, then its rate of cooling is given as
$$\begin{gathered} \frac{\mathrm{dQ}}{\mathrm{d}t}=K\left(T-T_0\right) \\ =\frac{2}{3}\left(30-20\right) \\ =\frac{20}{3}\mathrm{W} \end{gathered}$$
The initial rate of cooling when the body^,s temperature is 20°C is given as

$$\begin{gathered} {\left(\frac{\mathrm{dQ}}{\mathrm{d}t}\right)}_{20}=0 \\ \therefore {\left(\frac{\mathrm{dQ}}{\mathrm{d}t}\right)}_{30}=\frac{20}{3} \end{gathered}$$

${\left(\frac{\mathrm{dQ}}{\mathrm{d}t}\right)}_{\mathrm{avg}}=\frac{10}{3}$
t = 5 min = 300 s
Heat liberated$=\frac{10}{3}\times 300=1000\mathrm{J}$
Net heat absorbed = Heat supplied − Heat Radiated
= 6000 − 1000 = 5000 J
(d) Net heat absorbed is used for raising the temperature of the body by 10°C.
∴ m S ∆T = 5000
$$\begin{gathered} \mathrm{S}=\frac{5000}{m\times ∆T} \\ =\frac{5000}{1\times 10} \\ =500\mathrm{J}{\mathrm{Kg}}^{-1}{\mathrm{C}}^{-1} \end{gathered}$$