A piece of copper metal is initially at 100 C. It is dropped into a coffee cup calorimeter containing 50.0 g of water at a temperature of 20 C. After stirring, the final temperature of both copper and water is 25 C Assuming no heat losses, and that the specific heat (capacity) of water is 4.18 J (gK), what is the heat capacity of the copper in J/K?
To find the heat capacity of copper in joules per Kelvin (J/K), we can use the principle of conservation of energy.
The heat lost by the copper (q1) will be equal to the heat gained by the water (q2). The equation for heat transfer is given by:
q = m * c * ΔT
Where:
q is the heat transferred
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature
We can calculate q1 for the copper:
q1 = m1 * c1 * ΔT1
= m1 * c1 * (final temperature - initial temperature)
Given the specific heat capacity of water (c2) and the mass of water (m2), we can calculate q2 for the water:
q2 = m2 * c2 * ΔT2
= m2 * c2 * (final temperature - initial temperature)
Since q1 = q2, we can equate the equations and solve for the heat capacity of the copper (c1):
m1 * c1 * (final temperature - initial temperature) = m2 * c2 * (final temperature - initial temperature)
Since the initial temperature of the copper is 100°C and its final temperature is 25°C:
m1 * c1 * (25 - 100) = m2 * c2 * (25 - 20)
Simplifying the equation, we have:
m1 * c1 * (-75) = m2 * c2 * 5
Since mass of copper (m1) and mass of water (m2) are not provided, we can assume that the mass of water is sufficient to absorb all the heat from the copper. This means that m1 * c1 = m2 * c2.
Therefore, the equation becomes:
m1 * c1 * (-75) = m1 * c2 * 5
Substituting the given value for the specific heat capacity of water (c2 = 4.18 J/gK):
m1 * c1 * (-75) = m1 * 4.18 * 5
Simplifying further:
m1 * c1 * (-75) = 20.9 * m1
Dividing both sides of the equation by m1:
c1 * (-75) = 20.9
Finally, solving for c1 (the heat capacity of copper):
c1 = -20.9 / -75
c1 = 0.279 J/gK
Therefore, the heat capacity of copper is 0.279 J/K.
To find the heat capacity of the copper, we can use the equation:
\(q_{\text{copper}} = m_{\text{copper}} \cdot C_{\text{copper}} \cdot \Delta T_{\text{copper}}\)
where:
\(q_{\text{copper}}\) is the heat gained or lost by the copper (in this case, it will be the heat lost by the copper since the temperature decreases),
\(m_{\text{copper}}\) is the mass of the copper,
\(C_{\text{copper}}\) is the specific heat capacity of copper, and
\(\Delta T_{\text{copper}}\) is the change in temperature of the copper.
Since no heat is lost to the surroundings, the heat lost by the copper must be gained by the water. So we can use the equation:
\(q_{\text{water}} = m_{\text{water}} \cdot C_{\text{water}} \cdot \Delta T_{\text{water}}\)
where:
\(q_{\text{water}}\) is the heat gained by the water (equal to the heat lost by the copper),
\(m_{\text{water}}\) is the mass of the water,
\(C_{\text{water}}\) is the specific heat capacity of water, and
\(\Delta T_{\text{water}}\) is the change in temperature of the water.
Since the heat lost by the copper is equal to the heat gained by the water, we can equate the two equations:
\(m_{\text{copper}} \cdot C_{\text{copper}} \cdot \Delta T_{\text{copper}} = m_{\text{water}} \cdot C_{\text{water}} \cdot \Delta T_{\text{water}}\)
Now, we can solve for the heat capacity of the copper, \(C_{\text{copper}}\). Plugging in the known values:
\(m_{\text{copper}} =\) mass of copper (unknown)
\(m_{\text{water}} = 50.0 \, \text{g}\)
\(C_{\text{water}} = 4.18 \, \text{J/gK}\)
\(\Delta T_{\text{copper}} = 100 \, \text{°C} - 25 \, \text{°C}\)
\(\Delta T_{\text{water}} = 25 \, \text{°C} - 20 \, \text{°C}\)
We can rearrange the equation to solve for \(C_{\text{copper}}\):
\(C_{\text{copper}} = \frac{{m_{\text{water}} \cdot C_{\text{water}} \cdot \Delta T_{\text{water}}}}{{m_{\text{copper}} \cdot \Delta T_{\text{copper}}}}\)
Substituting the known values:
\(C_{\text{copper}} = \frac{{50.0 \, \text{g} \cdot 4.18 \, \text{J/gK} \cdot (25 \, \text{°C} - 20 \, \text{°C})}}{{m_{\text{copper}} \cdot (100 \, \text{°C} - 25 \, \text{°C})}}\)
Simplifying:
\(C_{\text{copper}} = \frac{{1045}}{{75 \cdot m_{\text{copper}}}}\)
Thus, the heat capacity of the copper is \(C_{\text{copper}} = \frac{{1045}}{{75 \cdot m_{\text{copper}}}}\) J/K.
[mass Cu x specific heat Cu x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Substitute and solve for specific heat Cu.