A chunk of copper weighing 18.9 grams and originally at 97.63oC is dropped into an insulated cup containing 77.9 grams of water at 22.44oC.
Assuming that all of the heat is transferred to the water, the final temperature of the water is what in C.
[mass water x specific heat water x (Tfinal-Tinitial)] + [mass copper x specific heat copper x (Tfinal-Tinitial) = 0
Solve for Tfinal.
To determine the final temperature of the water after the copper is dropped in, we can use the principle of heat transfer known as the heat gained equals heat lost.
The heat gained by the water can be calculated using the formula:
Q = m * c * ΔT
Where:
Q = heat gained by water
m = mass of water
c = specific heat capacity of water
ΔT = change in temperature of the water
The heat lost by the copper can be calculated using the same formula:
Q = m * c * ΔT
Where:
Q = heat lost by copper
m = mass of copper
c = specific heat capacity of copper
ΔT = change in temperature of the copper
Since the two substances are in thermal equilibrium, the heat gained by the water equals the heat lost by the copper:
m_water * c_water * ΔT_water = m_copper * c_copper * ΔT_copper
We can rearrange this equation to solve for the final temperature of the water:
ΔT_water = (m_copper * c_copper * ΔT_copper) / (m_water * c_water)
Now, substituting the given values:
m_water = 77.9 grams
c_water = 4.18 J/g°C (specific heat capacity of water)
ΔT_copper = -97.63°C (change in temperature of the copper)
m_copper = 18.9 grams
c_copper = 0.39 J/g°C (specific heat capacity of copper)
ΔT_water = (18.9 * 0.39 * -97.63) / (77.9 * 4.18)
Calculating this expression will give us the change in temperature of the water. We can then add this change to the initial temperature of the water (22.44°C) to find the final temperature of the water.