Question

A sample of stainless steel was stored in a freezer with a temperature of -4.0°C. After removing it from the freezer, the aluminum is placed in 1982 mL of water at room temperature (25.0°C). After several minutes thermal equilibrium is achieved at 21.4°C. What is the mass of the stainless steel sample? Please show all work with appropriate units.

Answer 1

First we assume that none of the heat transfer that takes place between steel and water is lost to the surroundings.

Since water is at higher temperature than refrigerated steel, so water loses heat and steel gains heat.

Heat loss/gain is calculated using the relation:

Q = m*C*dT

Here, m = mass, C = specific heat, dT = temperature

For water, C = 4.18 (J/g.C)

For steel, C = 0.48 (J/g.C)

Mass of water = Density*volume = (1 g/mL)*1982 mL = 1982 g

Heat lost by water, Qw = m*C*dT = 1982*4.18*(25-21.4) = 29825.14 J

Heat gained by steel, Qs = m*0.48*(21.4-(-4)) = m*12.192

Equating the above two expressions:

Heat gained Qs = Heat lost Qw

m*12.192 = 29825.14

So,

Mass of steel sample, m = 2446.29 g