Question

10 g of ice at 0°C is mixed with 100 g of water at 50°C in a calorimeter. The final temperature of the mixture is [Specific heat of water = 1 cal${g^{- 1}}^{o}C^{- 1}$ latent heat of fusion of ice = 80 cal $g^{- 1}$ ]

• A. 31.2°C
• B. 32.8°C
• C. 36.7°C
• D. 38.2°C

Answer 1

Answer: (D)

38.2°C

Answer 2

Here,

Mass of water, $m_{w} = 100g$

Mass of ice, $m_{i} = 10g$

Specific heat of water,

$s_{w} = 1calg^{- 1}{^\circ}C^{- 1}$

Latent heat of fusion of ice,

$L_{fi} = 80calg^{- 1}$

Let T be the final temperature of the mixture,

Amount of heat lost by water

$= m_{w}S_{w}(\Delta T)_{w} = 100 \times 1 \times (50 - T)$

Amount of heat gained by ice

$= m_{i}L_{fi} + m_{i}s_{W}(\Delta T)_{i} = 10 \times 80 + 10 \times 1 \times (T - 0)$

According to principle of calorimetry

Heat lost = Heat gained

$100 \times 1 \times (50 - T) = 10 \times 80 + 10 \times 1 \times (T - 0)$

500 – 10T = 80 + T

11T = 420 or T = $38.2{^\circ}C$