物理：比熱容計算題

1. Let T be the final temperature
Heat released by 2.8 g of steam
= [2.8 x 2 x (105-100) + 2.8 x 2260 + 2.8 x 4.2 x (100 -T)] J
= (7532 - 12T) J
where 2 J/kg.C is the specific heat capacity of steam, 2260 J/g is the latent heat of vapourization of water, and 4.2 J/g is the specific heat capacity of water.

Heat absorbed by ice
= . [12.3 x 2.1 x 4 + 12.3 x 334 + 12.3 x 4.2 x T] J
= (4212 + 52T) J

Heat absorbed by 1.7 g of water = 1.7 x 4.2 x (100-20) J = 571 J
Hence, using the heat balance equation,
(7532 - 12T) = (4212 + 52T) + 571
T = 43'C

2. Let T be the final temperature
Heat released by metal A = 26.5 x 263 x (82.6-T)
Heat absorbed by metal B = 10.3 x 189 x (T - 14.9)
Heat absorbed by water = 60 x 4200 x (T - 30)

Hence, 26.5 x 263 x (82.6-T) = 10.3 x 189 x (T - 14.9) + 60 x 4200 x (T - 30)
solve for T gives T = 31.3'C





2013-01-23 14:13:23 補充：
sorry, I made a mistake in Q1 in calculating the heat absorbed by water. Please replace the equation by the following:
Heat absorbed by 1.7 g of water = 1.7 x 4.2 x (T-20) J = (7.14T - 143) J

2013-01-23 14:14:26 補充：
Hence, using the heat balance equation,
(7532 - 12T) = (4212 + 52T) + (7.14T - 143)
T = 48.7'C