2.) A 48 g sample of water at 75 °C is mixed with 32 g water at 18 °C. Find the temperature of the mixture.
heat lost by 48 g + heat gained by 32 g = 0
[mass H2O x specific heat H2O x (Tfinal- Tinitial)] + [mass H2O x specific heat water x (Tfinal-Tinitial)= 0
Solve for Tfinal which is the only unknown in the equation.
To find the temperature of the mixture, we can use the principle of the conservation of energy.
First, let's calculate the amount of heat gained or lost by each sample of water using the specific heat capacity formula:
Q = m * c * ΔT
Where:
Q is the amount of heat gained or lost
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature
For the first sample of water:
m1 = 48 g
c1 = specific heat capacity of water (4.18 J/g°C)
ΔT1 = final temperature - initial temperature = Tf - 75°C
For the second sample of water:
m2 = 32 g
c2 = specific heat capacity of water (4.18 J/g°C)
ΔT2 = final temperature - initial temperature = Tf - 18°C
Since heat lost by one sample is equal to the heat gained by the other sample (assuming no heat loss to the surroundings), we can set up an equation:
m1 * c1 * ΔT1 = −(m2 * c2 * ΔT2)
Plugging in the values we know:
48 g * 4.18 J/g°C * (Tf - 75°C) = -(32 g * 4.18 J/g°C * (Tf - 18°C))
Now we can solve for Tf, the final temperature of the mixture.