Calculating specific heat capacity

Question

Assignment:

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 47.22 g of water at 19.7 °C. After closing and stiring the calorimeter thoroughly, the metal and water both come to equilibrium at a temperature of 27.6 °C.

Q1. What is the temperature change of the water?

• 100.0 °C
• 19.7 °C
• 7.9 °C

Q2. What is the temperature change of the metal?

• -72.4 °C
• 125.2 °C
• -27.6 °C

Q3. How much heat was gained by the water? (calculate the qwater)

• 33.1 J
• 1561 J
• 418.4 J

Q4. Knowing the above, what must qmetal?

• -33.1 J
• -1561 J
• -125240 J

Q5. Then what must the the Specific Heat of the metal be?

• 0.4184 J/g°C
• 25.00 J/g°C
• 0.1721 J/g°C