A block of ice of mass M = 0.6 kg, T_0 = 0degree is placed in a bucket containin

Question

A block of ice of mass M = 0.6 kg, T_0 = 0degree is placed in a bucket containing V_0 = 10 I of liquid water at T = 44degree C. Find the final temperature after the mixture comes to equilibrium. A block of ice of mass M = 0.6 kg, T_0 = 0degree is placed in a bucket containing V_0 = 2 l of liquid water at T = 4 degreeC. Find the remaining mass of ice after the mixture comes to equilibrium. A red hod block of steel (m = 1.0 kg, T_0 = 1150 K) is placed in a bucket containging boiling hot water to cool. Assume there is plenty of water to cool the steel down to the water's temperature. What mass of water is boiled off by the cooling steel?

Explanation / Answer

a)

density of water = 1000 kg/m^3

volume of water = 10 l = 10*10^(-3) m^3 = 10^(-2) m^3

mass of water = 10^(-2)*1000 = 10 kg

heat of fusion of ice = 333.55 kJ/kg

specific heat of water = 4.186 kJ/kg.C

equilibrium condition

0.6*333.55 + 0.6*4.186*T = 10*4.186*(44 - T)

200.13 + 2.5116^T = 1841.84 - 41.86*T

44.3716*T = 1641.71

T = 37 C

b)

let m kg is remaining mass of ice

volume of water = 2 l = 2*10^(-3) m^3

mass of water = 2*10^(-3)*1000 = 2 kg

equilibrium condition

(0.6-m)*333.55 = 2*4.186*4

m = 0.6 - 0.1004 = 0.4996 kg

c)

let the mass of water to boiled off is m.

specific heat of steel = 0.49 kJ/kg.C

1*0.49*1150 = m*4.186*100 + m*2257

m = 563.5/2675.6 = 0.21 kg = 210 gram water is boiled off

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