Finding the specific heat of a metal sample using calorimetry

Question

You perform a coffee-cup calorimetry experiment using a 6.694 g metal sample. You heated the metal sample to 99.5 degrees celcius as the initial temperature with a Bunsen burner before adding it to the coffee-cup. The initial temperature of the water in the calorimeter is 21.2 degrees celcius, the mass of the water in the calorimeter is 80.770 g, and the specific heat of water is 4.184 J/g*degrees celcius.  After you add the heated metal sample to the calorimeter, the calorimeter's final temperature is 24.6 degrees celcius, what was the specific heat of the metal sample? Make sure to attach a copy of your work!

Answer 1

The law of conservation of energy comes into play. The heat LOST by the hot metal, must be equal to the heat GAINED by the cooler water. The equation for heat lost or gained is...

q = mc∆T

For the water, we have...

q = heat gained = ?

m = mass of water = 80.770 g

c = specific heat = 4.184 J/gº

∆T = change in temperature = 24.6º - 21.2º = 3.4º

Solving for q (heat gained), we have ...

q = (80.770 g)(4.184 J/gº)(3.40º) = 1149 J heat gained by water

Recall that this MUST EQUAL the heat lost by the metal.

For the metal, we have ...

q = mc∆T

q = heat lost = 1140 J

m = mass = 6.694 g

c = specific heat = ?

∆T = change in temperature = 99.5º - 24.6º = 74.9º

Solving for c, we have ...

1140 J = (6.694 g)(c)(74.9º)

c = 2.27 J/gº