Question

Assume you use calorimetry to calculate the specific heat capacity of a 125.24 g piece of unknown metal. You intially heat the metal to 100.0 °C in boiling water. You then drop the chunk of metal into a calorimeter containing 45.22 g of water at 21.6 °C. After closing and stiring the calorimeter thoroughly, the metal and water both come to equilibrium at a temperature of 28.3 °C.

1. What is the temperature change of the water?

	6.7 °C
	21.6 °C
	100.0 °C

2. What is the temperature change of the metal?

	-71.7 °C
	-28.3 °C
	125.2 °C

3. How much heat was gained by the water? (calculate the q_water)

1268 J

418.4 J

28.0 J

4. Knowing the above, what must q_metal?

	-125240 J
	-28.0 J
	-1268 J

5. Then what must the the Specific Heat of the metal be?

0.1412 J/g°C

25.00 J/g°C

0.4184 J/g°C

Answer 1

The amount of heat lost or gained, q by a substance is given by

q = m × C × ∆T

where m = mass of substance in grams

C = heat capacity of the substance per g

∆T = temperature change