Question

2. A 237 g piece of molybdenum, initially at 100.0 °C, was dropped into 244 g of

Answer 1

Answer:

$\large\boxed{C = 0.270J/g\textdegree C}$

Explanation:

1. Energy balance

By the first law of thermodynamics, considering the system is closed and isolated, the heat released by the 237 g piece of molybdenum equals the heat absorbed by the 244 g of water.

2. Heat equation

The heat released or absorbed by a substance is proportional to the product of the mass, the specific heat and the change in temperature

• Q = m × C × ΔT

3. Heat released by the 237 g piece of molybdenum

At equilibrium:

• Q₁ = 237 g × C × (100.0ºC - 15.3ºC)

4. Heat absorbed by the 244 g of water

At equilibrium:

• Q₂ = 244 g × 4.184 J/gºC × (15.3ºC - 10.0ºC)

5. Solve for C from Q₁ = Q₂

• 237 g × C × (100.0ºC - 15.3ºC) = 244 g × 4.184 J/gºC × (15.3ºC - 10.0ºC)

• 20,073.9 × C = 5,410.7488 J/gºC

• C = 0.2695 J/gºC

The result must be reported with 3 significant figures: C = 0.270 J/gºC.