Question

In: Physics

a 50 gram ice cube at -15 deg centigrade is added to 400g of water at...

a 50 gram ice cube at -15 deg centigrade is added to 400g of water at 40 deg centigrade in an insulated cup of negligible heat capacity. what is the final equilibrium temp of the system? the heat of fusion and specific heat of ice are 80 cal/g and 0.5 cal/g deg centigrade. the specific heat of water is 1 cal/g deg centigrade.

Expert Solution

Specific heat of water = C₁ = 1 cal/(g.^oC)

Specific heat of ice = C₂ = 0.5 cal/(g.^oC)

Latent heat of fusion of ice = L = 80 cal/g

Mass of water = m₁ = 400 g

Mass of ice = m₂ = 50 g

Initial temperature of water = T₁ = 40 ^oC

Initial temperature of ice = T₂ = - 15 ^oC

Melting point of ice = T₃ = 0 ^oC

Final equilibrium temperature = T₄

The heat lost by the water is equal to the heat gained by the ice.

m₁C₁(T₁ - T₄) = m₂C₂(T₃ - T₂) + m₂L + m₂C₁(T₄ - T₃)

(400)(1)(40 - T₄) = (50)(0.5)(0 - (-15)) + (50)(80) + (50)(1)(T₄ - 0)

16000 - 400T₄ = 375 + 4000 + 50T₄

450T₄ = 11625

T₄ = 25.83 ^oC

Final equilibrium temperature of the system = 25.83 ^oC

genius_generous answered 2 years ago

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