Question

Four 2 cm × 2 cm × 2 cm cubes of ice are taken out from a refrigerator and are put in 200 ml of a drink at 10°C. (a) Find the temperature of the drink when thermal equilibrium is attained in it. (b) If the ice cubes do not melt completely, find the amount melted. Assume that no heat is lost to the outside of the drink and that the container has negligible heat capacity. Density of ice = 900 kg m−3, density of the drink = 1000 kg m−3, specific heat capacity of the drink = 4200 J kg−1 K−1, latent heat of fusion of ice = 3.4 × 105 J kg−1.

Answer 1

(I). The thermal equilibrium will be attained at 0°C.

(II). The mass of melted ice is 0.0247 kg.

Explanation:

Given that,

Number of ice cubes = 4

Volume of each ice cube = 8 cm³3

Density of ice = 900 kg m³

Total mass of ice, $m_{i} = (4\times8 \times10^{-6} \times900)$

$m_{i} = 288\times10^{-4}\ kg$

Latent heat of fusion of ice, $L_{i} =3.4 \times10^{5}\ J kg^{-1}$

Density of the drink = 1000 kg m⁻³

Volume of the drink = 200 ml

Mass of the drink $M= (200\times10^{-6})\times1000\ kg$

Let us first check the heat released when temperature of 200 ml changes from 10°C to 0°C

(a). We need to calculate the temperature of the drink when thermal equilibrium is attained in it

$H_{w}=M_{d}\times\rho_{d}\times C_{d}\times(T_{f}-T_{i})$

$H_{w}=200\times10^{-6}\times1000\times4200\times(10−0)$

$H_{w}=8400\ J$

Heat required to change four 8 cm³ ice cubes into water

$H_{i}=m_{i}L_{i}$

$H_{i}=288\times10^{-4}\times3.4 \times10^{5}$

$H_{i}=9792\ J$

Since, $H_{i} >H_{w}$,

The heat required for melting the four cubes of the ice is greater than the heat released by water

Some ice will remain solid and there will be equilibrium between  ice and water.

Thus, The thermal equilibrium will be attained at 0°C.

(b). We need to calculate the amount melted

Equilibrium temperature of the cube and the drink = 0°C

Let M be the mass of melted ice.

We need to calculate the heat released when temperature of 200 ml changes from 10°C to 0°C

Using formula of heat

$H_{w}=M_{d}\times\rho_{d}\times C_{d}\times(T_{f}-T_{i})$

$ML_{i}=H_{w}$

$ML_{i}=M_{d}\times\rho_{d}\times C_{d}\times(T_{f}-T_{i})$

Put the value into the formula

$M\times3.4 \times10^{5}=200\times10^{-6}\times1000\times4200\times(10−0)$

$M=\dfrac{8400}{3.4\times10^{5}}$

$M=0.0247\ Kg$

Hence, (I). The thermal equilibrium will be attained at 0°C.

(II). The mass of melted ice is 0.0247 kg.

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Topic : Mass of malted ice

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