Question

10) A 50 gram sample of a material requires 660 J of heat to have its temperature

Answer 1

Specific heat of the material thus calculated is 0.22 J/ g °C

Explanation:

Given:

Mass of the sample (m) = 50g

Heat energy = Q = 660 J

Temperature get raised from 20° C to 80° C

To Find:

Specific Heat of the material.

Formula to be used:

Q = m×C×ΔT

where we know that,

Heat energy = Q = 660 J

Mass = m = 50 g

ΔT = 80 - 20 = 60° C

Now to calculate the specific heat of the material (c) we must substitute all the values, Then we get,

$C=\frac{Q}{m \times \Delta T}$

  $=\frac{660 \mathrm{J}}{50 \mathrm{g} \times 60^{\circ} \mathrm{C}}$

  $=0.22 \frac{J}{g^{\circ} \mathrm{C}}$

Thus the Specific heat of the material is 0.22 J/ g °C.