When energy is added to a substance, its temperature generally increases. Likewise, when energy is removed from a substance, its temperature decreases. The amount of temperature change produced by a given amount of energy depends on both the amount of material present and on the nature of the substance itself.
Different substances respond differently to the addition of energy. For example, metals typically experience large temperature changes when relatively small amounts of energy are added, whereas water undergoes much smaller temperature changes for the same amount of energy. This behavior is described by a property called the specific heat, which measures the amount of energy required to raise the temperature of one gram of a substance by one degree Celsius.
The relationship between energy flow and temperature change is
\[ q = mC\Delta T \]
where \(q\) is the amount of energy transferred, \(m\) is the mass of the sample, \(C\) is the specific heat capacity, and \(\Delta T\) is the temperature change. This equation allows us to predict how much energy is required to heat or cool a substance, or alternatively how much the temperature will change when a known amount of energy is added or removed.
Each variable in the specific heat equation has a clear physical meaning:
| Symbol | Meaning | Typical Units |
|---|---|---|
| \(q\) | Energy transferred | J or kJ |
| \(m\) | Mass of the sample | g |
| \(C\) | Specific heat capacity | \(\mathrm{J\,g^{-1}\,^\circ C^{-1}}\) |
| \(\Delta T\) | Temperature change | \(^\circ\mathrm{C}\) or K |
The temperature change is always calculated as the final temperature minus the initial temperature:
\[ \Delta T = T_f - T_i \]
A positive value of \(q\) indicates that energy is absorbed by the substance, resulting in an increase in temperature. A negative value of \(q\) indicates that energy leaves the substance, resulting in a decrease in temperature.
Historically, the calorie was defined using water because water has a large and easily measured heat capacity. One calorie was originally defined as the amount of energy required to raise the temperature of one gram of water by one degree Celsius.
In modern SI units, energy is measured in joules, and the specific heat of liquid water is approximately
\[ C_{\mathrm{water}} = 4.184\ \mathrm{J\,g^{-1}\,^\circ C^{-1}} \]
This relatively large value means that water can absorb or release substantial amounts of energy while undergoing only modest temperature changes. This property helps regulate the temperature of lakes, oceans, and even living organisms.
Because water's specific heat is known so accurately, it is often used as a reference substance when studying heat flow and energy transfer.
A \(10.5\ \mathrm{g}\) sample of iron has a specific heat of \(0.442\ \mathrm{J\,g^{-1}\,^\circ C^{-1}}\). If its temperature is increased from \(25.0^\circ\mathrm{C}\) to \(100.0^\circ\mathrm{C}\), calculate the amount of energy absorbed by the iron.
The relationship between energy flow and temperature change is
\[ q = mC\Delta T \]
First calculate the temperature change:
\[ \Delta T = T_f - T_i \]
\[ \Delta T = 100.0^\circ\mathrm{C} - 25.0^\circ\mathrm{C} \]
\[ \Delta T = 75.0^\circ\mathrm{C} \]
Now substitute the values into the specific heat equation:
\[ q = \left(10.5\ \mathrm{g}\right) \left(0.442\ \frac{\mathrm{J}} {\mathrm{g\,^\circ C}}\right) \left(75.0^\circ\mathrm{C}\right) \]
Notice that the units of grams and degrees Celsius cancel:
\[ q = \left(10.5\right) \left(0.442\right) \left(75.0\right) \mathrm{J} \]
\[ q = 347.9\ \mathrm{J} \]
Rounding to three significant figures gives
\[ q = 348\ \mathrm{J} \]
Therefore, the iron absorbs \(348\ \mathrm{J}\) of energy as its temperature increases from \(25.0^\circ\mathrm{C}\) to \(100.0^\circ\mathrm{C}\).
Physical interpretation: Because the temperature increases, the value of \(q\) is positive. A positive value of \(q\) indicates that energy flows into the sample.
A \(50.0\ \mathrm{g}\) sample of water is initially at \(19.0^\circ\mathrm{C}\). If the water absorbs \(125.0\ \mathrm{J}\) of energy, calculate its final temperature.
Begin with the specific heat equation:
\[ q = mC\Delta T \]
Since the final temperature is unknown, solve the equation for \(\Delta T\):
\[ \Delta T = \frac{q}{mC} \]
The values given are
\[ q = 125.0\ \mathrm{J} \]
\[ m = 50.0\ \mathrm{g} \]
\[ C_{\mathrm{water}} = 4.184\ \frac{\mathrm{J}} {\mathrm{g\,^\circ C}} \]
Substitute these values into the equation:
\[ \Delta T = \frac{ 125.0\ \mathrm{J} }{ \left(50.0\ \mathrm{g}\right) \left( 4.184\ \frac{\mathrm{J}} {\mathrm{g\,^\circ C}} \right) } \]
Notice that the units of grams and joules cancel, leaving a temperature change in degrees Celsius:
\[ \Delta T = 0.5977^\circ\mathrm{C} \]
Since the water absorbs energy, its temperature increases:
\[ T_f = T_i + \Delta T \]
\[ T_f = 19.0^\circ\mathrm{C} + 0.5977^\circ\mathrm{C} \]
\[ T_f = 19.6^\circ\mathrm{C} \]
Therefore, the final temperature of the water is \(19.6^\circ\mathrm{C}\).
Physical interpretation: Water has a relatively large specific heat, so even though \(125.0\ \mathrm{J}\) of energy is added, the temperature increases by less than one degree Celsius.
An \(8.92\ \mathrm{g}\) sample of aluminum absorbs \(722\ \mathrm{J}\) of energy, causing its temperature to increase by \(90.3^\circ\mathrm{C}\). Calculate the specific heat of the metal.
Begin with the specific heat equation:
\[ q = mC\Delta T \]
Since the specific heat is unknown, solve the equation for \(C\):
\[ C = \frac{q}{m\Delta T} \]
The values given are
\[ q = 722\ \mathrm{J} \]
\[ m = 8.92\ \mathrm{g} \]
\[ \Delta T = 90.3^\circ\mathrm{C} \]
Substitute these values into the equation:
\[ C = \frac{ 722\ \mathrm{J} }{ \left(8.92\ \mathrm{g}\right) \left(90.3^\circ\mathrm{C}\right) } \]
Evaluating the expression gives
\[ C = 0.8967 \frac{\mathrm{J}} {\mathrm{g\,^\circ C}} \]
Rounding to three significant figures,
\[ C = 0.897 \frac{\mathrm{J}} {\mathrm{g\,^\circ C}} \]
Therefore, the specific heat of the metal is \(0.897\ \mathrm{J\,g^{-1}\,^\circ C^{-1}}\).
Physical interpretation: Specific heat is an intensive property of a substance. By measuring the energy absorbed, the mass of the sample, and the resulting temperature change, we can identify a material or compare its thermal properties to those of other substances.
Use \(q=mC\Delta T\) to solve the randomly generated problem below. Enter only the numerical value.
Big picture: Specific heat connects energy flow and temperature change. By knowing any three of the four quantities \(q\), \(m\), \(C\), and \(\Delta T\), the fourth can be determined. This relationship forms the foundation for calorimetry and the study of energy transfer in chemical and physical systems.