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

Question

stich3 [128]

4 years ago

7

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?

Chemistry

1 answer:

forsale [732] · Answer 1 · 2021-11-12T11:09:10+03:00

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)