Question

A 2.52g lead weight, initially at 10.1?C, is submerged in 7.94g of water at 52.3?C in an insulated container.

What is the final temperature of both substances at thermal equilibrium?

Answer 1

Specific heat of water = 4.184 J/g^oC

Specific heat of lead = 0.160 J/g^oC

At thermal equilibrium let us consider the temperature of both the substance are t ^oC.

Here heat absoebed by lead = - (Heat released by water)

Heat absoebed by lead = 2.52g X 0.160 J/g^oC X (t - 10.1)^oC

Heat released by water = 7.94g X 4.184 J/g^oC X (t - 52.3)^oC

So, 2.52g X 0.160 J/g^oC X (t - 10.1)^oC = -[7.94g X 4.184 J/g^oC X (t - 52.3)^oC]

t = 51.79 ^oC

At thermal equilibrium let us consider the temperature of both the substance are 51.79 ^oC.