A 0.2 kg block of ice at 0°C is dropped carefully into a styrofoam calorimeter cup with an unknown mass of water at 20°C. When thermal equilibrium is reached, the final temperature is measured to be 5°C. What was the mass of water initially in the cup? (Specific latent heat of fusion of ice = 334000 J/kg, specific heat capacity of water = 4200 J/kg K)
Answer (1)
The ice has to melt.... then that water has to warm to 5° C. The calories for this comes from the water already in the calorimeter cooling down 15° from 20° C.
Melt ice : .2 kg * 334000 J / kg = 66800 J
Warm this water : .2 kg * 4200 J / (kg K) * 5 = **4200 J **
Cool calorimeter water 15 ° : ** x * 4200 * 15 **
SO:
66800 + 4200 = x * 4200*15
**x = 1.13 kg of water originally in calorimeter. **
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