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

Answer:

i) Power given by heater = 160 W

ii) Power radiated by block = 16 W

iii) Power radiated by block at 25 degree C = 8 W

iv) time for which heater is on = 5.19 s

Explanation:

As we know that heat capacity of the block is given as

$C = 80 J/^oC$

now we know that heat given to raise the temperature of the block is given as

$Q = ms\Delta T$

$Q = C\Delta T$

now rate of heat transferred to the block is given as

$\frac{dQ}{dt} = C\frac{dT}{dt}$

so we will have

$\frac{dQ}{dt} = 80(2) = 160 W$

Part b)

Power Radiated by the block just after heater is switched off

$P = C\frac{dT}{dt}$

$P = 80(0.2) = 16 W$

Part c)

Power radiated by the block when its temperature is 25 degree C

Now we can use Newton's law of cooling

so we will have

$\frac{dQ}{dt} = k(T - T_s)$

$16 = k(30 - 20)$

$k = 1.6$

now when block temperature is given as 25 degree C then we have

$\frac{dQ}{dt} = 1.6(25 - 20)$

$\frac{dQ}{dt} = 8 W$

Part d)

As we know that power given by the heater is

P = 160 W

now we can use energy equations

Energy given by heater = Heat absorbed by block + heat loss

$160 t = 80(30 - 20) + 8 t$

$154 t = 800$

$t = 5.19 s$

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Topic : Heat transfer

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Answer 2

Explanation:

Given data,

Metal block  Heat capacity, $s=80 \mathrm{J}^{\circ} \mathrm{C}^{-1}$

Metal block heat absorbent is,

$H=s \times \Delta T=80 \times(30-10)=80 \times 20=1600 \mathrm{J}$

Rate of increase of block temperature= $\left(\frac{d \theta}{d t}\right)=2^{\circ} C / s$

Rate of fall in block temperature= $\left(\frac{d \theta}{d t}\right)=-0.2^{\circ} \mathrm{C} / \mathrm{s}$

The negative sign shows a decrease in temperature over time.

(a) Energy is  = s (Δθ)

Power is equal to  Energy per unit time

Heater power =Heat capacity $\times\left(\frac{d \theta}{d t}\right)$

P = 80 × 2

P = 160 W

(b) Radiated Power  = Energy lost /unit time.

P' = Heat capacity $\times\left(\frac{d \theta}{d t}\right)$

Here, $\left(\frac{d \theta}{d t}\right)$ represents the rate of temperature decrease.

   P' = 80 × 0.2

   P' = 16 W

(c) 16 W Power is radiated when block temperature decreases from 30 ° C to 20 ° C.

$s \times\left(\frac{d \theta}{d t}\right)_{d e c}=K\left(\theta-\theta_{0}\right)$

$16=K(30-20)$

    $\mathrm{K}=1.6$

Using newton's law of cooling,

$s \frac{d \theta}{d t}=K\left(\theta-\theta_{0}\right)$

$s \frac{d \theta}{d t}=1.6(30-25)$

$s \frac{d \theta}{d t}=1.6 \times 5$

$s \frac{d \theta}{d t}=8 \mathrm{watt}$