Question

A worker pours 1.250kg of molten lead at a temperature of 327.3 ?C into 0.5050kg of water at a temperature of 75.00 ?C in an insulated bucket of negligible mass.

Assuming no heat loss to the surroundings, calculate the mass of lead and water remaining in the bucket when the materials have reached thermal equilibrium.

Express your answer using four significant figures.

Answer 1

Latent heat of fusion of lead = 23 kJ/kg

Latent heat of vapourization of water = 2260 kJ/kg

Specific heat capacity of water (s) = 4.182 kJ/kg

Specific heat capacity of lead (l) = 0.160 kJ/kg

Case 1:

Heat released by fusion of lead = mass of lead * latent heat of fusion

Q = 1.25*23 = 28.75 kJ

Heat gain by water to go from temperature of 75 Degrees to 100 degrees = mass * s * 25

= .505 * 4.182 * 25 = 52.79 kJ

So, since this is more than the heat that is released so lead after solidifying will loose more heat :

So. let T be the change in temperature of lead :

1.25*0.162*T = (52.79-28.75)

T = 118.71

So, temperature of lead = 327.3-118.71 = 208.583 degrees

This temperature is still higher than water (that is at 100 degrees)

So now water will start vapourising:

Let m be the mass of water vapourized :

m*2260 = heat gained by water

Maximum heat that can be released by lead so that its final temperature will reach 100 degrees.

1.25* 0.162* (208.583-100) = 21.988 kJ

m*2260 = 21.988

m = 0.0097 kg = 9.72 g

So final temperature of system will be 100 degrees with 9.72 g of steam, 0.495 kg of water and 1.25 kg of solid lead