challenge problems

1) Assuming that there is no loss of heat energy to the surroundings, then:
Heat loss by the hot object = Heat gain by the cool object
Therefore,
Heat loss by the water + aluminium calorimeter = mc△T = (0.298 x 4200 + 0.105 x 900) x 5 = 6730.5 J
For the heat gain by the ice cube, it has two stages:
i) rising temperature in ice state from -7.6°C to 0°C.
ii) melting from ice to water state without change in temperature.
iii) rising temperature in water state from 0°C to 15°C.
Suppose the mass of the ice cube is m kg, then total heat gain by it is:
2100 x 7.6 x m (Stage 1) + x m (Stage 2) + 4200 x 15 x m (Stage 3)
= 412960 x m
Using the above relations, we have:
412960 x m = 6730.5
So, m = 0.0163 kg (corr. to 3 d.p.)
2) Suppose the mass of the meteorite is M kg. And using the following data:
Specific heat capacity of iron: 450 J kg-1 K-1
Specific latent heat of fusion of iron: 247.5 kJ K-1
Then upon melting, total heat energy absorbed by the iron meteorite is given by:
450 x [1808 - (-126)] x M + 247500 x M
= 1178000 x M
For simplification, we assume that all its original K.E. has been converted to its internal (heat) energy upon entering the earth's atmosphere which yields an equation:
(1/2)Mv2 = 1178000 x M
v = 1495 m s-1
Hence, we can see that the speed required is independent of its mass.
Note: For this question, emphasis should be again put on the assumption of total conversion of K.E. into the internal energy. In fact, for the meteorite, it can gain extra heat energy from its gravitational P.E. in the earth's gravitational field and the work done against the air in the atmosphere in falling down to the ground of the earth.

真係好challenge
第1題個aluminum calorimeter既specific heat capacity又冇
第2題個iron既melting point又冇
恕我無能為力