Question

A metal ball of mass 1 kg is heated by means of a 20 W heater in a room at ${20}^0$ C. The temperature of the ball becomes steady at ${50}^0$ C. (a) Find the rate of loss of heat to the surrounding when the ball is at ${50}^0$ C (b) Assuming Newton's law of cooling, calculate the rate of loss of heat to the surrounding when the ball is at ${30}^0$ C. (c) Assume that the temperature of the ball risks uniformly from ${20}^0$ C to ${30}^0$ 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 conditions no heat is absorbed. The rate of loss of heat by conduction is equal to that of the supplied.

m = 1 Kg,

Power of Heater = 20 W,

Room temperature = ${20}^0$ C

(a) H = $\frac{d\theta }{dt}$ = P = 20 watt

(b) By Newton's law of cooling

$\frac{d\theta }{dt}=K(\theta -{\theta }^0)$

$\Rightarrow 20=K(50-20),K=\frac{2}{3}$

Again, $\frac{d\theta }{dt}=K(\theta -{\theta }^0)$

= $\frac{2}{3}\times (30-20)=\frac{20}{3}$ W

(c) ${\left(\frac{d\theta }{dt}\right)}_{20}=0;{\left(\frac{d\theta }{dt}\right)}_{30}=\frac{20}{3}{\left(\frac{d\theta }{dt}\right)}_{avg}=\frac{10}{3}$

T =5 min = 300 s

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

Net Heat absorbed = Heat supplied -Heat Radiated

= 6000- 1000

= 5000 J

Now, $m\Delta \theta S$ = 5000

S = $\frac{5000}{m\Delta \theta }=\frac{5000}{1\times 10}$

$=500J-K{g}^{-1^{\circ }}{C}^{-1}$