An insulated container holds 4 kg of water at 20 Celsius . 0.5 kg of ice at -6 C

Question

An insulated container holds 4 kg of water at 20 Celsius . 0.5 kg of ice at -6 Celsius is added to the mixture . What is the final temp of the mixture?
Latent heat of fusion of ice is 33.5x10 ^4 j/kg
Specific of water at 20 degrees c is 4186 j/kg specific heat of ice 2093 j/kg An insulated container holds 4 kg of water at 20 Celsius . 0.5 kg of ice at -6 Celsius is added to the mixture . What is the final temp of the mixture?
Latent heat of fusion of ice is 33.5x10 ^4 j/kg
Specific of water at 20 degrees c is 4186 j/kg specific heat of ice 2093 j/kg
Latent heat of fusion of ice is 33.5x10 ^4 j/kg
Specific of water at 20 degrees c is 4186 j/kg specific heat of ice 2093 j/kg

Explanation / Answer

for ice from -6 C to 0 C

Q1 = mc*dT + m*Lf = 0.5*2093*6 + 0.5*33.5*10^4 = 1.737*10^5 J

now for ice from 0C to Tf

Q2 = 0.5*4186*Tf = 2093*Tf J

for water from 20 C to Tf C

Q3 = mc*dT = 4*4186*(20 - Tf) = 3.34*10^5 - 4*4186*Tf

Q1 + Q2 = Q3

1.737*10^5 + 2093*Tf = 3.34*10^5 - 4*4186*Tf

Tf = (3.34*10^5 - 1.737*10^5)/(2093 + 4*4186) = 8.509 C

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