Consider an aluminum calorimeter (m= 0.50 Kg) with 4.0 Kg of water. both the wat
ID: 3896750 • Letter: C
Question
Consider an aluminum calorimeter (m= 0.50 Kg) with 4.0 Kg of water. both the water and thecalorimeter begin at a temperature of 45 deg celcius. An unknown amount of ice (at 0 deg celcius) is added to the calorimeter until the temperature reaches 5.00 deg celcius.
a) how much heat does the original water in the calorimeter gain as it cools to 5.00 deg celcius?
b) how much heat does the ice absorb as it melts and warms to 5.00 deg celcius?
c) how much heat does the aluminun cup give up as it cools to 5.00 deg celcius?
d) how much ice is used to cool the calorimeter and water to a final temperature of 5.00 deg celcius?
e) how much steam at 125 deg celcius would you need to bring this system back to 45 deg celcius after it has been cooled by the ice?
Explanation / Answer
a) heat lost by water = m*C * delta T = 4 * 4.18 *(45-5) = 668.8kJ
heat lost by calorimeter(aluminium) = m*C* delta T = 0.5 * 0.897 *40 = 17.94kJ
b) heat gained by water = heat lost by calorimeter + heat lost by water = 686.74kJ
c) heat lost by aluminium cup = m*C* delta T = 0.5 * 0.897 *40 = 17.94kJ
d) let ice required be m,
m*h_lv + m*C*delta T = 686.74kJ
m*334 + m * 2.1 * 5= 686.74
m = 1.993 kg
e) Let m kg of steam be required
Heat released while cooling to 100 C = m*1.996*(125-100) = 49.9*m
Latent heat of vaporization = m*2260
Heat released while cooling to 45 C = m*4.18*(100 - 45) = 229.9*m
49.9*m + 229.9* m +2260 * m = 686.74
m = 0.27 kg