An electrically heating coil was placed in a calorimeter containing 360 gm of water at $10^{\circ} C$ . The coil consumes energy at the rate of 90 watt. The water equivalent of the calorimeter and the coil is 40 gm. Calculate, what will be the temperature of the water after 10 minutes.
(J = 4.2 Joule/cal.)

{22.14}^{\circ} C

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{42.14}^{\circ} C

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{142.14}^{\circ} C

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{420.14}^{\circ} C

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Solution

The correct option is B ${42.14}^{\circ} C$
Energy supplied by the heater to the system in 10 minutes:
$Q_{1} = P \times t = (90 J / s) \times (10 \times 60 s) = 54 k J$
$Q_{1} = (\frac{54}{4.2}) k c a l = 12857 c a l$
[as 4.2 J = 1 cal]
Now if T is the final temperature of the system, energy required to change its temperature from $10^{\circ} C$ to $T^{\circ} C$ is
i.e., $Q_{2} = (m + W)_{c} Δ T$
$(360 + 40) \times 1 \times (T - 10) c a l$
According to given problems
$Q_{1} = Q_{2} i . e . 400 (T - 10) = 12857$
or $T = {42.14}^{\circ} C$ .

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An electrically heating coil was placed in a calorimeter containing 360 gm of water at 10∘C. The coil consumes energy at the rate of 90 watt. The water equivalent of the calorimeter and the coil is 40 gm. Calculate, what will be the temperature of the water after 10 minutes. (J = 4.2 Joule/cal.)