Q.34 The temperature of a 12.58 g sample of calcium carbonate increases from 23.6 °C to 38.2C. If the specific heat of calcium carbonate is 0.82 J/g-K, how many joules of heat are absorbed?

150.6 Joules

Explanation:

Calorimetry problem must be solved by the calorimetry formula.

Q = c . m . ΔT

ΔT = Final T° - Initial T°

c = specific heat (Notice, that in the unit we have K as unit for T°, and the T° is in °C. We don't need to make the conversion, because is a Δ)

For example 100°C - 50°C = 50°C

373K - 323K = 50 K

So let's replace data given: Q = 0.82 J/g.K . 12.58 g . (38.2 - 23.6)

Q = 150.6 Joules

We assume no heat lost by sourrounding