The temperature of a 12.58 g sample of calcium carbonate [CaCO₃ (s)] increases from 23.6 °C to 38.2 °C. If the specific heat of calcium carbonate is 0.82 J/g-K, how many joules of heat are absorbed?

Question

Quizplus · Accepted Answer

The heat absorbed (q) can be calculated using the formula q = m * c * ΔT, where m is the mass (12.58 g), c is the specific heat (0.82 J/g-K), and ΔT is the change in temperature (38.2 °C - 23.6 °C = 14.6 °C). Thus, q = 12.58 g * 0.82 J/g-K * 14.6 K = 150.6 J, which rounds to 151 J.

The Temperature of a 12