A current of 2.50 A passing through a heating coil immersed in 180 g of paraffin (specific heat capacity 2.00 J g − 1 K − 1 ) contained in a 100 g calorimeter (specific heat capacity 0.400 J g − 1 K −

Question

A current of 2.50 A passing through a heating coil immersed in 180 g of paraffin (specific heat capacity 2.00 J g − 1 K − 1 ) contained in a 100 g calorimeter (specific heat capacity 0.400 J g − 1 K − 1 ) raises the temperature from 5 ∘ C below room temperature to 5 ∘ C above room temperature in 100 s . The reading of the voltmeter connected across the heating coil is

TestHub · Accepted Answer

The heat supplied by the coil is used to raise the temperature of the paraffin and the calorimeter.

Heat gained, $Q = (m_{p} c_{p} + m_{c} c_{c}) Δ T$

$Q = (180 g \times 2.00 J g^{- 1} K^{- 1} + 100 g \times 0.400 J g^{- 1} K^{- 1}) \times 10 K$

$Q = (360 J K^{- 1} + 40 J K^{- 1}) \times 10 K = 4000 J$

Power, $P = \frac{Q}{t} = \frac{4000 J}{100 s} = 40 W$

Also, $P = V I$ .

So, $V = \frac{P}{I} = \frac{40 W}{2.50 A} = 16.0 V$

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