A 60-g piece of ice at 0 °C is placed in 200 g of water that's in a 100-g aluminum calorimeter. The initial temperature of the water and calorimeter is 30 °C. In the following, use the accepted value for the latent heat of fusion of ice. (Keep lots of significant figures in your calculations below, since they are all intermediate results in a final calculation.) a. Find the final temperature when thermal equilibrium is reached. b. How much heat left the water and the calorimeter? c. How much of that heat went into melting the ice? d. How much of that heat went into warming up the melted ice? e. Final calculation: Confirm that the accounting of heat works out.

a) 6.60°C

b) 21,696.48 J = 21.70 kJ

c) 20,040 J = 20.04 kJ

d) 1,657.66 J = 1.66 kJ

e) Check Explanation

Explanation:

At thermal equilibrium,

Heat gained by the ice to melt and rise in temperature = Heat lost by water and the aluminium calorimeter

Mass of Ice = 60 g

Initial Temperature of ice = 0°C

Latent Heat of fusion of ice to water = 334 J/g

Initial Temperature of the water and aluminium = 30°C

Mass of water in calorimeter = 200 g

Mass of Aluminium calorimeter = 100 g

Specific heat capacity of water = 4.186 J/g.°C

Specific heat capacity of Aluminium = 0.90 J/g.°C

Let the final temperature of the setup = T

Heat gained by ice to melt at 0°C = mL = (60×334) = 20,040 J

Heat gained by the melted ice to rise in temperature from 0°C to final temperature T

= mCΔT = (60×4.186) × (T - 0) = (251.16T) J

Heat lost by water in the calorimeter = mC ΔT

= (200×4.186) × (30 - T) = (25,116 - 837.2T) J

Heat lost by the aluminium calorimeter = mCΔT = (100×0.90) (30 - T) = (2700 - 90T) J

a) At thermal equilibrium,

Heat gained by the ice to melt and rise in temperature = Heat lost by water and the aluminium calorimeter

20,040 + 251.16T = 25,116 - 837.2T + 2700 - 90T

251.16 T + 837.2 T + 90 T = 25,116 + 2700 - 20,040

1,178.36 T = 7,776

T = (7,776/1178.36) = 6.60°C

b) How much heat left the water and the calorimeter?

Heat lost by the water and calorimeter

= (25,116 - 837.2T) + (2700 - 90T)

T = 6.60°C

25,116 - (837.2×6.6) + 2700 - (90×6.6) = 21,696.48 J

c. How much of that heat went into melting the ice?

This is already calculated, heat gained by the ice to melt the ice = mL = (60×334) = 20,040 J

d. How much of that heat went into warming up the melted ice?

Heat gained by the melted ice to rise in temperature from 0°C to final temperature T

= mCΔT = (60×4.186) × (T - 0) = (251.16T) J

T = 6.60°C

251.16T = (251.16 × 6.6) = 1,657.656 J

e. Final calculation: Confirm that the accounting of heat works out

Heat gained by the ice to melt and rise in temperature = Heat lost by water and the aluminium calorimeter

20.04 + 1.66 = 21.70

21.70 = 21.70

Hence, the accounting of heat works!

Hope this Helps!!!

54 degrees incident angle = reflected angle

angle is from normal turned to line of ray