Question

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

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?

Homework Answers

Answer #1

Specific heat of water = 4.184 J/goC

Specific heat of lead = 0.160 J/goC

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/goC X (t - 10.1)oC

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

So, 2.52g X 0.160 J/goC X (t - 10.1)oC = -[7.94g X 4.184 J/goC 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.

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