Ice of mass 2kg at -20°C and 5 kg of water at 20°C are mixed (specific heat of water = 1 Cal/g°C ,specific heat of ice = 0.5 Cal/g°C).Find the net amount of water in the container.

Question

Abhishek Kumar · Answer

Let the mass of ice converted to water be m. So, 1000mL = (5+m)1000 x 1 x (T-293)

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Ice of mass 2kg at -20°C and 5 kg of water at 20°C are mixed (specific heat of water = 1 Cal/g°C ,specific heat of ice = 0.5 Cal/g°C).Find the net amount of water in the container.

Ice of mass 2kg at -20°C and 5 kg of water at 20°C are mixed (specific heat of water = 1 Cal/g°C ,specific heat of ice = 0.5 Cal/g°C).Find the net amount of water in the container.

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Ice of mass 2kg at -20°C and 5 kg of water at 20°C are mixed (specific heat of water = 1 Cal/g°C ,specific heat of ice = 0.5 Cal/g°C).Find the net amount of water in the container.

Ice of mass 2kg at -20°C and 5 kg of water at 20°C are mixed (specific heat of water = 1 Cal/g°C ,specific heat of ice = 0.5 Cal/g°C).Find the net amount of water in the container.

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