heat lost by Cu + heat gained by H2O = 0
[mass Cu x specific heat Cu x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
[mass Cu x specific heat Cu x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
The amount of heat gained by the water = The amount of heat lost by the copper.
The heat gained or lost by an object can be calculated using the formula:
Q = m × c × ΔT
Where:
Q is the heat gained or lost
m is the mass
c is the specific heat capacity
ΔT is the change in temperature
In this case, the water gains heat and the copper loses heat. We can calculate the heat gained by the water and the heat lost by the copper separately and equate them.
For the water:
M1 = Mass of water = 93.4 g
C1 = Specific heat capacity of water = 4.18 J/g·°C
T1 = Initial temperature of water = 21.0°C
Tf = Final temperature = 24.2°C
Using the formula above, we can calculate the heat gained by the water as:
Q1 = M1 × C1 × (Tf - T1)
For the copper:
M2 = Mass of copper (unknown)
C2 = Specific heat capacity of copper = 0.39 J/g·°C
T2 = Initial temperature of copper = 69.1°C
Tf = Final temperature = 24.2°C
Using the formula above, we can calculate the heat lost by the copper as:
Q2 = M2 × C2 × (Tf - T2)
Since the total heat gained by the water is equal to the total heat lost by the copper, we can set up the equation:
Q1 = Q2
M1 × C1 × (Tf - T1) = M2 × C2 × (Tf - T2)
Now we can substitute the known values into the equation and solve for M2, the mass of copper:
93.4 g × 4.18 J/g·°C × (24.2°C - 21.0°C) = M2 × 0.39 J/g·°C × (24.2°C - 69.1°C)
Simplify and solve for M2:
(93.4 × 4.18 × 3.2) / (0.39 × (-44.9)) = M2
M2 ≈ 194.4 g
Therefore, the mass of the copper block is approximately 194.4 grams.