Using the relationship for heat gained or lost by a substance
Try these questions to test your ability to use the relationship for specific heat capacity.
Question
An electric heater supplies 13500 joules of heat energy to a metal block of mass \(0.5kg\). The temperature of the block rises from \(20^{\circ}C\) to \(80^{\circ}C\) during the heating process. Assuming that very little heat is lost from the block during the heating process, what is the specific heat capacity of the metal?
In order to calculate the specific heat capacity of the metal you need to take the following steps:
\(m=0\cdot5\,kg \)
\(\Delta T=\left ( 80-20 \right ) \)
\(=60^{o}C \)
\(Heat\,gained\,E_{h}=cm\Delta T \)
\(c=\frac{E_{h}}{m\Delta T} \)
\(\frac{13500}{\left ( 0\cdot5\times60 \right )}\)
\(=\,450\)
Specific heat capacity of the metal = \(450 J kg^{-1^{\circ}}C^{-1}\)
Question
Calculate the energy is required to raise the temperature of \(750g\) of water from \(34^{\circ}C\) to \(91^{\circ}C\).
Use the Data Sheet to obtain the value for the specific heat capacity for water (on page 2 of Section 1 of the exam paper).
\(c = 4180 J kg^{-1^{\circ}}C^{-1}\), calculate temperature change \(\Delta T =(91-34)+57^{\circ}C\) convert \(750g\) into\(0.75 kg\).
\(E_{h}=mc\Delta T \)
\(E_{h}=0\cdot75\,\times\,4180\,\times57 \)
\( E_{h}= 178695\,J\)
Heat energy required = \(178695 J\)
Question
When \(6.3\times10^{5} J \) of heat energy is added to a statue in a kiln oven, the statue’s temperature rises from \(16^{\circ}C\) to \(95^{\circ}C\). The specific heat capacity of the statue material is \(3500 J kg^{-1^{\circ}}C^{-1}\).
Calculate the mass of the statue.
\(E_{h}\,=\,6.3\,\times\,10^{5}J, \Delta T = (95-16) = 79^{\circ}C, c = 3500 J kg^{-1^{\circ}}C^{-1}\)
\(E_{h}=mc\Delta T \)
\(6.3\times10^{5}=m\times 3500\times79\)
\(m=\frac{6\cdot3\,\times10^{5}}{3500\,\times79} \)
\(m=2.28\,kg\)
Question
Calculate the final temperature of \(120g\) of water when \(2·4\times 10^{4}J\) of heat energy is added to it. The initial temperature of the water is \(19^{\circ}C\).
\(E_{h}= 2\cdot4 \times10^{4}J\)
\(E_{h}=mc\Delta T \)
\(2\cdot4\,\times10^{4}=0\cdot12\,\times4180\,\times\Delta T \)
\(\Delta T=\frac{2\cdot4\times10^{4}}{0\cdot12\times4180} \)
\(\Delta T=47\cdot9\,^{o}C\)
The final temperature of the water is \(19 + 47.9 = 66.9^{\circ}C\)
