HEAT AND THERMAL ENERGY: Specific Heat Capacity Determination

Q: A quantity of lead shots weighing 100 g are placed in a beaker full of boiling water. 

After the shots reach a temperature of 100.0 ℃, they are transferred to a vessel that contains 100 g of water at 20.0 ℃. After sitting for some time, the water cools to an equilibrium temperature of 22.4 ℃. What is the specific heat capacity of the lead?

A: The specific heat capacity of the lead shots is determined using the following equation: 

Q = mc_sΔT_s

c_s = ( Q / mΔT_s )

The mass of the shots is known, so we must determine how much of a decrease in temperature occurred within them, and we must ascertain how much heat energy was transferred from one system to the other:

ΔT_s = ( T_sf – T_si )

ΔT_s = ( 22.4 ℃ – 100.0 ℃ )

ΔT_s = – 77.6 ℃

The heat energy that was lost by the shots was gained by the water:

– Q lost = + Q gained

The specific heat capacity of water is 4.186 kJ / kg*K, and there are 100 g of water present; therefore, the specific heat of the water system is obtained via multiplication:

( Specific Heat Capacity )( Mass ) = Specific Heat

Note: Be sure to use appropriate units at all times!!! The quantity of water is expressed in grams ( g ), but the version of specific heat capacity here has its mass expressed in kilograms ( kg ). Fortunately, we can reduce the kilogram and kilojoule ( kJ ) terms of the numerator and denominator by 1,000. This will establish consistency of units between the constant used and the units within which water has been measured:

( 4.186 J / g*K )( 100 g ) = 418.6 J / K

Take careful note of how much less the temperature of the water changed in comparison to that of the lead!!! This perfectly demonstrates how different materials and mediums accommodate heat energy differently:

ΔT_w = ( T_wf – T_wi )

ΔT_w = ( 22.4 ℃ – 20.0 ℃ )

ΔT_w = + 2.4 ℃

Please recall that changes in temperature on the Celcius scale are directly proportional to temperature changes recorded using the Kelvin ( K ) scale. For this reason, we substitute a ( K ) in place of ( ℃ ) in our ΔT_w term:

( 418.6 J / K )( + 2.4 K ) = + 1,004.6 J

Since this is the amount of heat energy gained by the water, it is also the amount of heat energy that was lost by the lead:

– Q = – 1,004.6 J

We now have the information needed to determine the specific heat capacity of the lead shots; however, the version of specific heat capacity of interest to us is expressed in terms of kilograms, so we convert 100 g to 0.100 kg prior to substitution:

c_s = [ – 1,004.6 J / ( 0.100 kg )( – 77.6 K ) ]

c_s = 129.5 J / kg*K ≅ 0.1295 kJ / kg*K

The actual value of the specific heat capacity of lead is 0.13 kJ / kg*K, which is in excellent agreement with our experimental results!!!