The specific heat capacity of nickel is 0.440 J/(gK). If you have a nickel ball that starts at 100.0 °C and gets dropped into a calorimeter containing 50.0 g of water (yes, you should know the specific heat of water), which goes from 20.0 °C to 23.5 °C. What is the mass of the nickel ball?
[mass Ni x specific heat Ni x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal - Tinitial)] = 0
Substitute and solve for mass Ni which is the only unknown in the equation.
Post your work if you get stuck.
i came up with 733.74? Not sure if thats correct.
To calculate the mass of the nickel ball, we can use the principle of heat transfer. The heat lost by the nickel ball will be equal to the heat gained by the water. The equation used for this principle is:
\( q_{\text{nickel}} = q_{\text{water}} \)
To calculate the heat lost by the nickel ball (\( q_{\text{nickel}} \)), we can use the equation:
\( q_{\text{nickel}} = m_{\text{nickel}} \times c_{\text{nickel}} \times \Delta T_{\text{nickel}} \)
Where:
\( m_{\text{nickel}} \) is the mass of the nickel ball (what we want to find)
\( c_{\text{nickel}} \) is the specific heat capacity of nickel (given as 0.440 J/(gK))
\( \Delta T_{\text{nickel}} \) is the change in temperature of the nickel ball, which is initially at 100.0 °C and we assume reaches the final temperature of the water.
To calculate the heat gained by the water (\( q_{\text{water}} \)), we can use the equation:
\( q_{\text{water}} = m_{\text{water}} \times c_{\text{water}} \times \Delta T_{\text{water}} \)
Where:
\( m_{\text{water}} \) is the mass of the water (given as 50.0 g)
\( c_{\text{water}} \) is the specific heat capacity of water (which is approximately 4.18 J/(gK))
\( \Delta T_{\text{water}} \) is the change in temperature of the water, which goes from 20.0 °C to 23.5 °C.
Setting the two equations equal to each other, we have:
\( m_{\text{nickel}} \times c_{\text{nickel}} \times \Delta T_{\text{nickel}} = m_{\text{water}} \times c_{\text{water}} \times \Delta T_{\text{water}} \)
Plugging in the known values:
\( m_{\text{nickel}} \times 0.440 \times (23.5 - 100.0) = 50.0 \times 4.18 \times (23.5 - 20.0) \)
Now we can solve this equation to find the mass of the nickel ball (\( m_{\text{nickel}} \)).