Assume you use calorimetry to calculate the specific heat capacity of a 125.24 g

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?

2. What is the temperature change of the metal?

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

1268 J

418.4 J

28.0 J

4. Knowing the above, what must qmetal?

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

Explanation / Answer

q = (45.22 g) (4.184 J/g-1 °C-1) (6.7 °C)

q = 1268 J

4. calorimeter got to rest at q = 0 (At equilibrium)

i.e., heat gained by water + heat lost by metal = 0

Heat lost by metal = -1268 J

5. specific heat of metal can be calculated as:

q = m Cp T

-1268J= (125.24 g) Cp(-71.7 °C)

Cp= 0.1412 J/g°C

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.