#### Question

The temperature of 170g of water at 50°C is lowered to 5°C by adding a certain amount of ice to it. Find the mass of ice added.

#### Solution

Given mass of water = 170 g = 0.17 kg

T1 = 50°C, T2 = 5°C,

specific heat capacity of water = 4200 J kg –1°C–1

Sp. Latent heat of ice = 336000 J kg–1

Let mass of ice added be x then we have heat lost by water = Heat used by ice.

⇒ water at 50°C to 5°C = ice at °C to water at °C + water at 0°C to 5°C

`=> 0.17 xx 4200 xx (50 - 5)`

`= x xx 336000 + x xx 4200 xx 5 = 32130`

`= 357000 x`

or, `x = 32130/357000`

`= 0.09 Kg`

`x = 90 g`

we should add 90 g of ice to water.

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The Temperature of 170g of Water at 50°C is Lowered to 5°C by Adding a Certain Amount of Ice to It. Find the Mass of Ice Added. Concept: Calorimetry - Specific Heat Capacity.

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