Heat Capacity Formula

Heat Capacity Formula - Solved Examples & Practice Question

When heat is given to a body, its temperature increases and when the body loses heat, its temperature falls. Heat capacity is the heat required to raise the temperature of an object by one degree. It can be calculated as the ratio of the amount of heat energy given to an object to the resulting increase in its temperature.
Heat Capacity is expressed by,
\[c = \frac{{\Delta Q}}{{\Delta T}}\]
Wherein, ΔQ = amount of heat transferred, ΔT = increase in temperature.
(An equal fall in temperature is observed when the body loses the same amount of heat).
In SI units, heat capacity is expressed in joule per kelvin (J/K)
In Heat capacity, the mass of the body can have any value (not specified like unit mass etc.).

Example:
If 6400 J of heat supplied to a body raises its temperature by 100 degrees, then find its heat capacity.
Solution: \[\Delta Q = 6400\,J,\,\,\,\Delta T = 100^\circ ,\,\,\,c = ?\]
Heat capacity; \[c = \frac{{\Delta Q}}{{\Delta T}} = \frac{{6400}}{{100}} = 64\,J/K\,\,\]
Question:
Two samples of water A and B are given heat separately. For sample A, 12600 J of heat increases its temperature by 6 degrees. For B, 9450 J of heat increases its temperature by 4.5 degrees. Compare the heat capacities of A and B (consider ideal situations).
Options:
(a) cA > cB
(b) cA < cB
(c) cA = cB
(d) data insufficient.
Answer: (c)