物理問題，幫幫手

Be aware that the specific heat capacity of ice is not the same as water. For ice, the specific heat capacity is around 2100 J/kg-C, which is only half of that of water.

1. Let T be the final temperature, heat abosrbed by ice
= 0.1 x 2100 x 10 + 0.1 x 334000 + 0.1 x 4200 x T
= 35500 + 420T

Heat released by water
= 1.2 x 4200 x (30-T)
Use heat balance equation,
35500 + 420T = 1.2 x 4200 x (30-T)
solve for T gives T = 21.2'C

2. Here, the mass of ice is much larger than that of water. It may happen that the final temperature T is below the melting point of ice (i.e. 0'C).

Heat released by water
= 0.1 x 4200 x 10 + 0.1 x 334000 + 0.1 x 2100 x T
= 37600 + 210T
Heat absorbed by ice = 1 x 2100 x (0 - T) = - 2100T

Applying the heat balance equation,
37600 + 210T = -2100T
solve for T gives T = -16.3'C

第一題
(1.2)(4200)(30-T)=(0.1)(4200)(T-(-10))
T=26.9°C(cor. to 3sig.fig.)

第二題
(0.1)(4200)(10-T)=(1)(4200)(T-(-20)
T=-17.3°C(cor. to 3sig.fig.)


law of conservation of energy 好簡單，唔知點解你可以唔識做