Question

A metal block of heat capacity 80 J°C⁻¹ placed in a room at 20°C is heated electrically. The heater is switched off when the temperature reaches 30°C. The temperature of the block rises at the rate of 2°C s⁻¹ just after the heater is switched on and falls at the rte of 0.2 °C s⁻¹ just after the heater is switched off. Assume Newton's law of cooling to hold. (a) Find the power of the heater. (b) Find the power radiated by the block just after the heater is switched off. (c) Find the power radiated by the block when the temperature of the block is 25°C. (d) Assuming that the power radiated at 25°C represents the average value in the heating process, find the time for which the heater was kept on.

Answer 1

Given:
Heat capacity of the metal block, s = 80 J°C⁻¹
Heat absorb by the metal block is,
$H=s\times ∆T=80\times (30-10)=800\mathrm{J}$

$$\begin{gathered} \mathrm{Rate}\mathrm{of}\mathrm{rise}\mathrm{in}\mathrm{temperature}\mathrm{of}\mathrm{the}\mathrm{block}=\left(\frac{d\theta }{dt}\right)=2°\mathrm{C}/\mathrm{s} \\ \mathrm{Rate}\mathrm{of}\mathrm{fall}\mathrm{in}\mathrm{temperature}\mathrm{of}\mathrm{the}\mathrm{block}=\left(\frac{d\theta }{dt}\right)=-0.2°\mathrm{C}/\mathrm{s} \end{gathered}$$
The negative sign indicates that the temperature is falling with time.

(a) Energy = s ($∆\theta$)
Power = Energy per unit time

∴ Power of the heater$=\mathrm{Heat}\mathrm{capacity}\times \left(\frac{d\theta }{dt}\right)$
P = 80 × 2
P = 160 W

(b) Power Radiated = Energy lost per unit time.
$\therefore P\text{'}=\mathrm{Heat}\mathrm{capacity}\times \left(\frac{\mathit{d}\mathit{\theta }}{\mathit{d}\mathit{t}}\right)$
Here, $\frac{\mathit{d}\mathit{\theta }}{\mathit{d}\mathit{t}}$ represents the rate of decrease in temperature.
P' = 80 × 0.2
P' = 16 W

(c) 16 W of power is radiated when the temperature of the block decreases from 30°C to 20°C.
$\therefore s\times {\left(\frac{d\mathrm{\theta }}{dt}\right)}_{\mathrm{dec}}=K\mathit{}\left(\theta \mathit{-}{\theta }_{\mathit{0}}\right)$
⇒ 16 = K (30 − 20)
K = 1.6

From newton's law of cooling,
$$\begin{gathered} \therefore s\frac{d\theta }{dt}=K\left(\theta -{\theta }_0\right) \\ \Rightarrow s\frac{d\theta }{dt}=1.6\left(30-25\right) \\ \Rightarrow s\frac{d\theta }{dt}=1.6\times 5 \\ \Rightarrow s\frac{d\theta }{dt}=8\mathrm{Watt} \end{gathered}$$