Steam at 10 0 ∘ C is passed into 2.0 kg of water contained in a calorimeter of water equivalent 0.02 kg at 1 5 ∘ C till the temperature of the calorimeter and its contents rise to 9 0 ∘ C . The mass o

Question

Steam at 10 0 ∘ C is passed into 2.0 kg of water contained in a calorimeter of water equivalent 0.02 kg at 1 5 ∘ C till the temperature of the calorimeter and its contents rise to 9 0 ∘ C . The mass of steam condensed in kg is

TestHub · Accepted Answer

The principle of calorimetry states that heat lost equals heat gained.

Heat gained by water and calorimeter:

$Q_{g ain} = (m_{w} c_{w} + m_{c} c_{w}) (T_{2} - T_{1})$

$Q_{g ain} = (2.0 k g + 0.02 k g) \times 4200 J k g^{- 1} K^{- 1} \times (9 0^{\circ} C - 1 5^{\circ} C)$

$Q_{g ain} = 2.02 \times 4200 \times 75 = 636300 J$

Heat lost by steam:

$Q_{l oss} = m L_{v} + m c_{w} (T_{s} - T_{2})$

$Q_{l oss} = m (2.26 \times 1 0^{6} J k g^{- 1}) + m (4200 J k g^{- 1} K^{- 1}) (10 0^{\circ} C - 9 0^{\circ} C)$

$Q_{l oss} = m (2.26 \times 1 0^{6} + 42000) = m (2.302 \times 1 0^{6}) J k g^{- 1}$

Equating heat gained and lost:

$636300 = m (2.302 \times 1 0^{6})$

$m = \frac{636300}{2.302 \times 1 0 ^{6}} \approx 0.2764 k g$

The closest option is 0.280 kg.

The final answer is $0.280$ .

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