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`        A metal block of heat capacity 80 J/°C 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 just after the heater is switched on and falls at the rate of 0.2°C/s 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 time for which the heater was kept on.`
3 years ago

Jitender Pal
365 Points
```										Sol. Given:
Heat capacity = m × s = 80 J/°C
(dθ/dt) base increase = 2 °C/s
(dθ/dt) base decrease = 0.2 °C/s
(a) Power of heater = mS (dθ/dt) base increa sin g = 80 * 2 = 160 W
(b) Power radiated = mS (dθ/dt) base decrea sin g = 80 * 0.2 = 16 W
(c) Now mS (dθ/dt) base decrea sin g = K (T – T base 0)
⇒ 16 = K(30 - 20) ⇒ K = 16/10 = 1.6
Now dθ/dt = K(T – T base 0) = 1.6 * (30 - 25) = 1.6 * 5 = 8 W
(d) P.t = H ⇒ 8 * t

```
3 years ago
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