Question

A piece of ice of mass 40 g is added to 200 g of water at 50 °C. Calculate the final temperature of water when all the ice has melted. Specific heat capacity of water = 4200 J kg^-1K^-1, and specific latent heat of fusion of ice = 336 × 10³ J kg^-1.

Answer 1

Mass of ice = 40 gm = `40/1000 = 0.04` kg

Mass of water = 200 g = `200/1000` = 0.2 kg

Temperature = 50°C

Final temperature of water T = ?

Specific heat of water = 4200 J kg^-1K^-1

Specific latent heat of fusion of ice

= 336 ×10³ J kg^-1

Heat energy taken by ice to melt at 0°C

= mL

= 0.04 × 336 × 10³

 = 13440 J

Heat energy taken by ice to rise from 0°C to T°C

= mass × specific heat capacity × rise in temperature

= 0.04 × 4200 × T

= 168 T

Heat energy given by water in fall of its temperature from 50°C to T = m × c × change in temperature

= 0.2 × 4200 × (50 − T)

= 840 (50 − T)

= 42000 − 840 T

If there is no loss of heat.

Heat energy given by water = Heat energy taken by ice

⇒ 42000 − 840 T = 13440 + 168 T

⇒ 42000 − 13440 = 168 T + 840 T

⇒ 28560 = 1008 T

⇒ T = `28560/1008`

⇒ T = 28.330°