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·˚c, what is the mass (in grams) of the sample?

Given:

Q = 45.5 J, amount of heat absorbed

ΔT = 28.5 - 21.1 = 7.4 °C = 7.4 K, temperature change

c = 0.82 J / (g-°C), specific heat of CaCO₃.

Le m = the mass of the sample (g).

Then

Q = mcΔT

(45.5 J) = (m g) * (0.82 J / (g-°C)) * (7.4 °C)

m = 45.5 / (0.82*7.4) = 7.4984 g

Answer: 7.5 g (nearest tenth)