A single 50g ice cube is dropped into a thermally insulated container holding 20

Question

A single 50g ice cube is dropped into a thermally insulated container holding 200g of water. The water is initially at 25°C and the ice is initially at-15oC 1) what is the final temperature of the system after is has come to thermal equilibrium ? 2.49 Submit 2) In terms of mass, how much of the ice has melted? grams Submit 3) Now lets drop a second 50gm cube of ice into the system. What is the final temperature of the system after it has come to thermal equilibrium for this second time? Submit 4) What is the total mass of ice (including the first cube) melted during this entire process grams Submit

Explanation / Answer

heat required for one piece of ice to melt

Qreq = m*c*dt + m*Lf = (0.05*2030*15)+(0.05*334*10^3) = 18222.5 J

heat lost by water in melting ice = Qav = 0.2*4190*25 = 20950 J

here the toal ice melts

1)

Qgain = Q loss

18222.5 + 0.05*4190*t = 0.2*4190*(25-t)

t = 2.6 degrees

2)

mass = 50 grams

3)

heat required to melt the two ice pieces = 2*18222.5 = 36445 J

maximum heat available with water to melt = 20950 J

the second ice piece cannot melt completely

the system is in equilibrium at 0 degrees

4)

let m gram melts

m*c*dt + m*Lf = 20950

m*2030*15+m*335*10^3 = 20950

m = 0.0573 kg = 57.3 grams

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