50g of ice at 0 degree Celsius is mixed with 40g of water at 60 degree celsius then ratio of ice and water at equilibrium is?
Answers
At 0° C the the heat given by the water is :
ΔT = 60°C - 0°C = 60°C
Specific heat capacity of water is : 4186 J / kg
Heat given = mass of water × specific heat capacity of water × change in temperature
60 × 4186 × 0.04 Kg = 10046.4Joules
The heat gained by ice to melt = mLf where m is the mass of ice and Lf the latent heat of fusion of ice.
Lf = 334000J/kg
Thus heat required to melt the ice is thus :
0.05 × 334000 = 16700Joules.
As we can see the heat given by the hot water is less than the heat required to melt the ice implying that the heat will only melt a portion of the ice.
The mass of the ices melted is :
m × 334000 = 10046.4
m = 10046.4 / 334000 = 0.0300 Kg
The melted ice is thus : 0.03Kg
0.03Kg× 1000 = 30g
The ice remaining unmelted is :
50g - 30g = 20g
The total amount of water in liquid state is :
30g + 40g = 70g
The ratio of ice to water is thus :
20 / 70 = 2/7
2 : 7
The equilibrium temperature is 0°C since it is at this temperature that ice melts to water.
At 0° C the the heat given by the water is :
ΔT = 60°C - 0°C = 60°C
Specific heat capacity of water is : 4186 J / kg
Heat given = mass of water × specific heat capacity of water × change in temperature
60 × 4186 × 0.04 Kg = 10046.4Joules
The heat gained by ice to melt = mLf where m is the mass of ice and Lf the latent heat of fusion of ice.
Lf = 334000J/kg
Thus heat required to melt the ice is thus :
0.05 × 334000 = 16700Joules.
As we can see the heat given by the hot water is less than the heat required to melt the ice implying that the heat will only melt a portion of the ice.
The mass of the ices melted is :
m × 334000 = 10046.4
m = 10046.4 / 334000 = 0.0300 Kg
The melted ice is thus : 0.03Kg
0.03Kg× 1000 = 30g
The ice remaining unmelted is :
50g - 30g = 20g
The total amount of water in liquid state is :
30g + 40g = 70g
The ratio of ice to water is thus :
20 / 70 = 2/7
2 : 7