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

Explanation:

The body has achieved equilibrium in steady state. So there will be no more sharing of heat between the body and the surrounding.

This implies that in stable condition,

Heat loss rate = The rate at which heat is transmitted

Given data ,

Mass (m) = 1 kg

Power heater = 20 W

Temperature = 20°C

(a)At stable state,

Heat loss rate = The rate at which heat is transmitted

Loss rate / heat gain = Power

$\frac{d Q}{d t}=P=20 W$

(b) Through Newton's cooling theorem, the cooling rate is directly proportional to the temperature difference.

So if the body is in a steady state, its cooling rate is given as

$\frac{d Q}{d t}=K\left(T-T_{0}\right)$)

$20=K(50-20)$

$K=\frac{2}{3}$

(c)When the body temperature is 30 ° C, its cooling rate is given as

$\frac{d Q}{d t}=K\left(T-T_{0}\right)=\frac{2}{3}(30-20)=\frac{20}{3}$

The initial cooling rate is given when the temperature of the body is 20 ° C

$\left(\frac{d Q}{d t}\right)_{20}=0$

$\left(\frac{d Q}{d t}\right)_{30}=\frac{20}{3}$

$\left(\frac{d Q}{d t}\right)_{a v g}=\frac{10}{3}$

$t=5 \min =300 \mathrm{s}$

Liberated Heat= $\frac{10}{3} \times 300=1000$

Total absorbed heat = supplied Heat – Radiated Heat

                            $=6000-1000=5000 \mathrm{J}$

(d) total absorbed heat is used to increase body temperature by 10 ° C.

$\mathrm{m} \mathrm{S} \Delta \mathrm{T}=5000$

$S=\frac{5000}{m \times \Delta T}=\frac{5000}{1 \times 10}=500 \frac{J}{K g C}$