Question

A sample of rhodium is heated to 137.50°C. The sample is then placed in a cup containing 11.3g of water at 23.30°C. The final temperature of the water and rhodium is 26.41°C. Calculate the mass of the sample of rhodium.
The specific heat of water is 4.184J/g°C
The specific heat of rhodium is 0.242J/g°C

Answer 1

By conservation of total energy of the system we know that the heat released from cooling of heated rhodium solution is all used up in heating of the water.

Lets say we had x g of rhodium sample which was heated to 137.5 C and the final temperature after cooling is 26.41 C the amount of heat produced Q will be given by Q = mCT where m is the mass of the rhodium sample which is x g , C is the specific heat of rhodium = 0.242 J/g-C and T is the change in temperature of the sample which will be initial temperature - final temperature = 137.5 C - 26.41 C = 111.09 C

Applying the values we get Q_r = (x g)*(0.242 J/g-C)*(111.09 C) = 26.88378*x J

The heat consumed by the water can also be calculated similarly, mass of water = 11.3 g , specific heat of water = 4.184 J/g-C and T = final temperature - initial temperature = 26.41 C - 23.30 C = 3.11 C

So, applying values in Q = mCT we get Q_w = (11.3 g)*(4.184 J/g-C)*(3.11 C) = 147.038312 J

Equating heat evolved from rhodium to heat consumed by water we get

Qw = Qr

so, 26.88378*x J = 147.038312 J so, x = 147.038312 /26.88378 g = 5.4694 g