The temperature of a 11.8 g sample of calcium carbonate [CaCO₃ (s)] increases from 24.8 °C to 38.0 °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 can be calculated using the formula q = m*c*ΔT, where m is mass, c is specific heat, and ΔT is the change in temperature. Substituting the given values: q = 11.8 g * 0.82 J/g-K * (38.0 °C - 24.8 °C) = 130 J.

The Temperature of a 11