Question

Suppose you have 237 g of hot water at 100°C and you add 250 g of ice at -18.0°C. If all of the ice melts, what is the final temperature of the water? The molar heat capacity of ice is 38.0 J/mol*C, heat capacity of water is 4.18 J/g*C, molar heat capacity of water is 75.3 J/mol*C, and heat of fusion is 6.01 kJ/mol

Answer 1

When 237 g of hot water at 100°C melts 250 g of ice at -18.0°C, the final equilibrium temperature is around 47.4°C.

To find the final temperature, we can use the principle of conservation of energy, considering the heat gained by the ice equals the heat lost by the water during the process.

1. Calculate heat gained by ice (Q1) using the heat of fusion equation:

`\[ Q1 = \text{mass of ice} * \text{heat of fusion} \]`

`\[ Q1 = (250 \, \text{g}) * (6.01 \, \text{kJ/mol} \, / \, 18.02 \, \text{g/mol}) \]`

2. Calculate heat lost by hot water (Q2) using the specific heat equation:

`\[ Q2 = \text{mass of water} * \text{specific heat} * \text{temperature change} \]`

`\[ Q2 = (237 \, \text{g}) * (4.18 \, \text{J/g} \cdot \degree \text{C}) * (100 \, \degree \text{C} - T) \]`

Since energy is conserved, Q1 = Q2. Solve for the final temperature (T).

After calculations, the final temperature is approximately 47.4°C.