A 1-kg block of aluminum initially at 20 degree C is dropped into a large vessel

Question

A 1-kg block of aluminum initially at 20 degree C is dropped into a large vessel of liquid N_2 which is boiling at 77 K. If the vessel is thermally insulated from its surroundings, calculate the number of liters of N_2 that will boil off by the time the aluminum reaches 77 K. [Given: sp. heat of nitrogen = 0.21 cal/g. degree C heat of vaporization of nitrogen = 48 cal/g density of nitrogen = 0.8 g/cm^3 sp. heat of aluminum = 0.215 cal/g. degree C Determine the amount of ice at -10 degree C that needs to be added to 4.0 kg of water at 10 degree C to cause the resulting mixture of ice and water to reach thermal equilibrium at 5 degree C. You must assume no heat transfer to the surrounding environment, so that heat transfer occurs only between the water and ice. Given: Specific heat of ice = 2.05 times 10^3 J/kg.K Latent heat of fusion of ice = 3.335 times 10^5 J/kg Specific heat of water = 4186 J/kg.K A heavy copper pot of mass 2.0 kg (including copper lid) is at a temperature of 150 degree C. You pour 0.10 kg water at 25 degree C into the pot so that no steam can escape. Assuming that the final temperature of the copper pot & water reaching at 100 degree C, determine the fraction of water that changes to gaseous phase. Also find the amount of water which is converted into steam. Given: Specific heat of water C_water = 4190 J/kg.K; C_copper = 390 J/kg. K Laten heat of vaporization of water L_v = 2.256 times 10^6 J/kg.

Explanation / Answer

problem 3rd:

m(copper) = 2.0 kg

T(copper) = 150°C

c(copper) = 390 J/kg*K

m(water) = 0.10 kg

T(water) = 25°C

c(water) = 4190 J/kg*K

Lv(water) = 2256x10^3 J/kg

Heat transfer: Q = mc(delta)T

Heat transfer in Phase Change: Q = m*(Lv)

Conservation of energy: Q(pot) + Q(water) = 0

I applied the heat transfer equation Q = mc(delta)T for both the water and the copper pot. I assumed the final temperature to be 100°C and calculated both values separately. Since energy is conserved, I knew that Q(pot) + Q(water) = 0 but then I immediately saw there was a difference that did NOT equal 0. That difference is the energy the copper pot lost to convert water into steam! In my case it was -7575 J, then it was a matter of figuring out how much steam was converted using Q = m*Lv. Regardless of which Lv I used for water, the magic of sig figs resulted in a m = 0.0034 kg or 3.4g

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