A system receives heat continuously at the rate of 10 W- The temperature of the system becomes constant at 70- C when the temperature of the surrounding is 30- C- After the heat is switched off- the system cools from 50- C to 49-9- C in 1min- Find the heat capacity in kJ- C of the system-

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Standard V

Science

Airflow

A system rece...

Question

A system receives heat continuously at the rate of $10 W$ . The temperature of the system becomes constant at $70^{\circ} C$ when the temperature of the surrounding is $30^{\circ} C$ . After the heat is switched off, the system cools from $50^{\circ} C$ to ${49.9}^{\circ} C$ in $1 min$ . Find the heat capacity (in $k J /^{\circ} C)$ of the system.

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Solution

Formula used: $\frac{d Q}{d t} = k (Δ θ)$

At $70^{\circ} C$ , the system attains steady state.
i.e., Rate of heat generated = Rate of heat loss

$\frac{d Q}{d t} = k (Δ θ)$

$10 W = k (70 - 30)^{\circ} C$

$k = (\frac{1}{4}) W /^{\circ} C$

At $50^{\circ} C$ , rate of heat loss should be,

$\frac{d Q}{d t} = k (Δ θ) = k (50 - 30)^{\circ} C = \frac{1}{4} \times 20 = 5 W,$

$But rate of heat loss = (heat capacity) \times (rate of cooling)$

$\frac{- d Q}{d t} = c (\frac{- d T}{d t})$

$- 5 W = c {[\frac{49.9 - 50}{60}]}^{\circ} C / s$

$c = 3000 J /^{\circ} C$

$c = 3 k J /^{\circ} C$

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