2. 2 kg of water at 23 °C is placed in a 3 kW freezer. (SHC of water = 4200 J kg–1 K–1)
(i) How long would it take to convert the water to ice at −15 °C?
(ii) How much longer would it take if the freezer was only 85% efficient?
Solution
List the stages and the amount of heat energy involved in each:
23 °C water → 0 °C water: temperature change, Q = mc ∆q
0 °C water → 0 °C ice: state change, Q = ml
0 °C ice → −15 °C ice: temperature change, Q = mc ∆q
Total heat energy = (2)(4200)(23) + (2)(3.3 × 105) + (2)(2100)(15)
= 193.2 kJ + 6.6 × 105 J + 63 kJ = 916.2 kJ (i) Time to convert =
Energy required ____________________
Power = 916 200
___________ 3000 = 305.4 s
(ii) If 85% efficient, only 85% of power is available: 916 200
________________ (0.85)(3000) = 359.29 s
H
3. 50 g of iron at 120 °C is added to a 100 g copper calorimeter containing 120 g of water at 25 °C. Assuming that the calorimeter is at the same temperature as the water and there are no external heat losses, calculate the final temperature of the water.
(SHC of iron = 448 J kg−1 K −1, SHC of copper = 390 J kg−1 K −1, SHC of water = 4200 J kg−1 K−1)
You may now complete Exercise 11A (page 112). FIG. 11.4 Solution
Let t be the final temperature. We can state the rise and fall relative to t for each part.
The water and copper calorimeter will increase in temperature from 25 °C. Therefore the ∆q rise will be (t − 25).
The iron will decrease in temperature from 120 °C.
Therefore the ∆q fall will be (120 − t). State the transfer of heat equation:
Heat gained by water and calorimeter = Heat lost by added iron mwcw∆qw + mccc∆qc = mici∆qi
504(t − 25) + 39(t − 25) = 22.4(120 − t) 504t − 12 600 + 39t − 975 = 2688 − 22.4t 565.4t = 16 263 t = 28.76 °C The final temperature of the water is 28.76 °C.
