Heat Capacity, \(C\)
Heat capacity is the quantity of heat needed to raise the temperature of the object by \(1\ ^\circ C\).
\(C = \dfrac{Q}{\Delta\theta}\)
Where \(Q\) = quantity of heat supplied (\(J\)) and \(\Delta\theta\) = change in temperature (\(^\circ C\)).
SI unit: \(J\ ^\circ C^{-1}\).
Specific Heat Capacity, \(c\)
Specific heat capacity is the quantity of heat needed to raise the temperature of \(1\ kg\) mass of the substance by \(1\ ^\circ C\).
\(c = \dfrac{Q}{m\Delta\theta}\)
or the heat absorbed/released: \(Q = mc\Delta\theta\)
SI units: \(J\ kg^{-1}\ ^\circ C^{-1}\) or \(J\ kg^{-1}\ K^{-1}\).
Meaning of Value
The specific heat capacity of aluminium is \(900\ J\ kg^{-1}\ ^\circ C^{-1}\). This means \(1\ kg\) of aluminium requires \(900\ J\) of heat to raise its temperature by \(1\ ^\circ C\).
Key Difference
Heat capacity depends on the mass of the object as well as its type of material. Specific heat capacity depends only on the type of material.
Worked Example: Conservation of Thermal Energy (\(Q_1 = Q_2\))
\(20\text{ g}\) of boiling water at \(100\ ^\circ C\) is poured into a glass containing \(200\text{ g}\) of water at \(28\ ^\circ C\). Calculate the final temperature of the mixture.
Let \(y\) = final temperature of the mixture.
Water that releases heat: \(m_1 = 20\text{ g} = 0.02\text{ kg}\), \(\Delta\theta_1 = (100 - y)\ ^\circ C\)
Water that absorbs heat: \(m_2 = 200\text{ g} = 0.20\text{ kg}\), \(\Delta\theta_2 = (y - 28)\ ^\circ C\)
Specific heat capacity of water, \(c = 4\ 200\ J\ kg^{-1}\ ^\circ C^{-1}\).
\(Q_1 = Q_2 \;\Rightarrow\; m_1 c\Delta\theta_1 = m_2 c\Delta\theta_2\)
\(0.02(4\ 200)(100 - y) = 0.20(4\ 200)(y - 28)\)
\(8\ 400 - 84y = 840y - 23\ 520\)
\(924y = 31\ 920\)
\(y = 34.55\ ^\circ C\)
Therefore, the final temperature of the mixture is \(34.55\ ^\circ C\).
Assumption: No heat is absorbed from or released to the surroundings. Heat transfer occurs only between the boiling water and the water at \(28\ ^\circ C\), so the heat released by the boiling water equals the heat absorbed by the water at \(28\ ^\circ C\).
Watch the temperature differences: the body that releases heat gives \((100 - y)\) and the body that absorbs heat gives \((y - 28)\), so both changes stay positive.