a) In an adiabatic calorimeter containing a mass of water m1=1980 g at the temperature T1=21 °C, we add a mass of water m2=360 g at the temperature T2=60 °C. Calculate the equilibrium temperature if we neglect the heat capacity of the calorimeter.
Question
a) In an adiabatic calorimeter containing a mass of water m1=1980 g at the temperature T1=21 °C, we add a mass of water m2=360 g at the temperature T2=60 °C. Calculate the equilibrium temperature if we neglect the heat capacity of the calorimeter.
Solution
To solve this problem, we need to use the principle of conservation of energy. This principle states that the total amount of energy in an isolated system remains constant. In this case, the energy lost by the hot water must be equal to the energy gained by the cold water.
The formula for the heat gained or lost by a substance is given by Q = mcΔT, where:
- Q is the heat energy,
- m is the mass of the substance,
- c is the specific heat capacity of the substance, and
- ΔT is the change in temperature of the substance.
For water, the specific heat capacity c is approximately 4.18 J/(g°C).
Let's denote the final equilibrium temperature as Tf. The heat lost by the hot water is then Q_hot = m2c(T2 - Tf), and the heat gained by the cold water is Q_cold = m1c(Tf - T1).
Setting these two quantities equal to each other gives us the equation:
m2c(T2 - Tf) = m1c(Tf - T1)
We can simplify this equation by dividing both sides by c, and then solve for Tf:
m2(T2 - Tf) = m1(Tf - T1)
Rearranging terms gives:
m2T2 - m2Tf = m1Tf - m1T1
Solving for Tf gives:
Tf = (m2T2 + m1T1) / (m1 + m2)
Substituting the given values into this equation gives:
Tf = (360g * 60°C + 1980g * 21°C) / (1980g + 360g)
Calculating this gives:
Tf ≈ 26.4°C
So, the final equilibrium temperature of the water in the calorimeter is approximately 26.4°C.
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