First slide

Specific heat

Question

Water of volume 2 L in a closed container is heated with a coil of 1 kW. While water is heated, the container loses energy at a rate of 160 J/s. In how much time the temperature of water rise from 27°C to 77°C? (Specific heat of water is 4.2 kJ/kg and that of the container is negligible).

Moderate

A

8 min 20 s
B

6 min 2 s
C

7 min
D

14 min

Solution

Mass of water is $ρ = \frac{m}{v}$
$m = ρv = 10^{3} \times 2 \times 10^{- 3} = 2 kg$ (since 1000 L = 1m³)
Now, using $ΔQ = msΔT = (p - p') \times t$
$= 2 \times 4 .2 \times 10^{3} (77 - 27) = (1000 - 160) t$
$\Rightarrow t = \frac{2 \times 4 .2 \times 10^{3} \times 50}{840} = \frac{500}{60} s = 8 \min 20 s .$

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