Carl adds a 6.70-gram cube of ice at 0.0°C to cool a glass of juice, As the ice melts, the temperature of the newly melted water rises to 25.4°C. Water’s heat of fusion is 334 joules/gram, and its specific heat is 4.186 joules/gram degree Celsius. How many joules of energy did the ice cube gain?

2,950

joules.

Explanation:

The units don’t need conversion, and the formula doesn’t need changing.

First, determine the energy required to melt the ice:

Q = mHf

Q = 6.70 g × 334

Q = 2,240 J

Next, determine the energy needed to raise the temperature of the newly melted water:

Q = mCΔT

Q = 6.70 g × 4.186 × (25.4°C – 0.0°C)

Q = 712 J

Now, add the energy values. Express the answer to 3 significant figures:

2,240 J + 712 J = 2,950 J.