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

answer: the rate of diffusion increases with increase in temperature.

increase in temperature, increases the molecular/particles speed (kinetic energy. diffusion means movement of particles through the material.

generally, the heavier the particle the more it interacts with the environment and more the temperature matters. so higher temperature results in higher diffusion.