# Specific Heat of Ice and Calorimetry

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A 40 g block of ice is cooled to -78 degrees C and is then added to 560 g of water in an 80 g copper calorimeter at a temperature of 25 degrees C

Determine the final temperature of the system consisting of the ice, water and calorimeter. If all the ice melts, determine how much ice is left.

Remeber that the ice must first warm to 0 degrees C, melt and then continue warming as water.

The specific het of ice is 0.500cal/gram C = 2090J/kg C

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##### Solution Summary

The solution calculates the temperature of a system consisting of a block of ice, water and a copper calorimeter.

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A 40 g block of ice is cooled to -78 degrees C and is then added to 560 g of water in an 80 g copper calorimeter at a temperature of 25 degrees C

Determine the final temperature of the system consisting of the ice, water and calorimeter. If all the ice melts, determine how much ice is left.

Remeber that the ice must first warm to 0 degrees C, melt and then continue warming as water.

The specific het of ice is 0.500cal/gram C = 2090J/kg C

specific heat capacity of

Copper= 386 J/Kg/C or 0.386 J/g/C

Water= 4,180 J/Kg/C or 4.18 J/g/C

Ice= 2,090 J/Kg/C or 2.09 J/g/C

Latent heat of fusion of ice = 334,000 J/Kg or 334 J/g

Heat = mass x specific heat capacity x change in temperature

Step 1: Calculate the heat required to melt all ice:

Heat required to change the temperature of ice from

-78 to 0 degrres C ie by 78 degrees C

mass of ice= 40 g

specific heat capacity of ice= 2.09 J/g/C

change in tempearture= 78 degrees C

Therefore, heat required= 6,520.80 J =40 x 2.09 x 78

Heat required to melt ice

heat required= mass x latent heat of fusion of ice

mass of ...

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