A sample of calcium carbonate [CaCO₃ (s)] absorbs 45.5 J of heat, upon which the temperature of the sample increases from 21.1 °C to 28.5 °C. If the specific heat of calcium carbonate is 0.82 J/g-K, what is the mass (in grams)of the sample?

Question

Quizplus · Accepted Answer

We can use the formula for heat (q) to solve for the mass (m) of the sample:

q = mCΔT

where q = 45.5 J, C = 0.82 J/g-K, ΔT = (28.5-21.1) = 7.4 K

Solving for m:

m = q / (CΔT) = 45.5 J / (0.82 J/g-K x 7.4 K)

m = 7.5 g

Therefore, the mass of the sample is 7.5 grams.

The answer is C only.

A Sample of Calcium Carbonate [CaCO3 (S)] Absorbs 45

A Sample of Calcium Carbonate [CaCO₃ (S)] Absorbs 45