A sample of calcium carbonate [CaCO₃ (s)] absorbs 40.3 J of heat, upon which the temperature of the sample increases from 20.8 °C to 27.3 °C. If the specific heat of calcium carbonate is 0.82 J/g-K, what is the mass (in grams)of the sample? A)7.6 g B)5.1 g C)-7.6 g D)0.13 g E)5.3 g

Question

Quizplus · Accepted Answer

Using the formula q = mcΔT, where q is the heat absorbed (40.3 J), m is the mass, c is the specific heat (0.82 J/g-K), and ΔT is the change in temperature (27.3 °C - 20.8 °C = 6.5 °C), we can solve for m: m = q / (cΔT) = 40.3 J / (0.82 J/g-K * 6.5 K) = 7.6 g.

A Sample of Calcium Carbonate [CaCO3 (S)] Absorbs 40

A Sample of Calcium Carbonate [CaCO₃ (S)] Absorbs 40