Ice, Water, Steam A block of ice of mass M = 0.6 kg, T_ o = 0 degree is placed i

Question

Ice, Water, Steam A block of ice of mass M = 0.6 kg, T_ o = 0 degree is placed in a bucket containing V_ o = 10 l of liquid water at T = 44 degree C. Find the final temperature after the mixture comes to equilibrium. A block of ice of mass M = 0.6 kg, T_o = 0 degree is placed in a bucket containing V_ o = 2 l of liquid water at T = 4 degree C. Find the remaining mass of ice after the mixture comes to equilibrium. A red hod block of steel (m = 1.0 kg, T_o = 1150 K) is placed in a bucket containing 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

V0 = 10 x 10^-3 m^3

mass of water = 1000x 10 x 10^-3 = 10 kg

water releases heat and absorbing heat, ice convert into water and then temp increases.

Using energy conservation,

ML + M Cwater deltaT = m Cwater deltaT

0.6 x 334 +   (0.6 x 4.186 x (T - 0 )) = 10 x 4.186 x (44 - T)

200.4 + 2.5116T = 1841.84 - 41.86T

T = 37 deg C

b) mass of water = 2 kg

heat released by water as temp decreases to 0 deg C,

Q = 2 x 4.186 x (4 - 0 )   = 33.488 kJ

this heat will use to change the state of ice.

Q = mL

33.488 = m x 334

m = 0.1 kg

remaining mass of ice = 0.6 - 0.1= 0.5 kg

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