What would be the final temperature if a 2kg piece of lead at 250∘C is inserted in a container with a 15kg of water at 40∘C? (for lead, c= 128 J/kg∘C)

Question

I solved this and got 208.68 (not rounded) did I solve this correct? Because others got 34.8

Answer 1

Heat lost by lead = heat gained by water

The specific heat for lead should be 128 kJ/kg/deg, not 128 J, or change it to 0.128 J/g/deg.

q = mc∆T and q lead = q water so (mc∆T lead) = (mc∆T water)

Let Tf = final temperature

(2 kg)(128 kJ/kg/deg)(250 deg - Tf) = (15 kg)(4184 kJ/kg/deg)(40 deg - Tf)

64,000 - 256Tf = 62,760Tf - 2,510,400

2,574,400 = 63,016Tf

Tf = 40.85 degrees C = final temperature of the water.

This makes sense, because 15 kg of water started at 40º and 2 kg plead (small relatively to 15 kg) was placed in the water.  Of course, the final temperature of the water CANNOT be 34.8ºC because that is LESS than the initial temperature. Heat from the lead, will flow into the water thus RAISING the temperature of the water.