Question

A 4.2 kg lead ball is dropped into a 2.4 L insulated pail of water initially at 20.0∘C.

If the final temperature of the water-lead combination is 26.0 ∘C, what was the initial temperature of the lead ball?

Answer 1

First you need to know (or look up) the specific heat capacity of lead, which is 129 J / (kg K), and of water, which is 4181.3 J / (kg K). The specific heat equation tells us that the amount of heat (Q) added is given by the equation:

Q = cmΔT

Since the heat added to the water is equal to the heat taken from the lead, we know that

CMΔT = -cmΔt

where C, M, and ΔT are the water's specific heat capacity, mass, and temperature change, and c, m, and Δt are those for lead. Solving for Δt gives:

Δt = ΔT(CM)/(cm)

We know that:

ΔT = 26 °C - 20.0 °C = 4 °C = 4 K (temperature difference, not absolute temperature)
C = 4181.3 J / (kg K)
M = 2.40 L (1 kg/L) = 2.4 kg
c = 129 J / (kg K)
m = 4.2 kg

Inserting these numbers into the above equation yields:

Δt = (4 K) [4181.3 J / (kg K)] (2.4 kg) / [129 J / (kg K)] (4.2 kg)
= 74.08 K (= 74.08° C)

So the lead ball was initially at 26° C + 74.08° C = 100.08° C