150g of ice at 0°C is mixed with 300g of water at 50°C. Calculate the temperature of the mixture
To calculate the temperature of the mixture, we can use the principle of conservation of energy.
The energy gained by the ice as it warms up to the final temperature is equal to the energy lost by the water as it cools down to the final temperature.
The energy gained or lost by a substance can be calculated using the specific heat capacity formula:
Q = mcΔT
Where:
Q is the energy gained or lost (in joules)
m is the mass of the substance (in grams)
c is the specific heat capacity of the substance (in J/g°C)
ΔT is the change in temperature (in °C)
For the ice:
m = 150g
c = 2.09 J/g°C (specific heat capacity of ice)
ΔT = Tf - 0°C, where Tf is the final temperature
For the water:
m = 300g
c = 4.18 J/g°C (specific heat capacity of water)
ΔT = 50°C - Tf, where Tf is the final temperature
Since the energy gained by the ice is equal to the energy lost by the water, we can set up the equation:
Qice = Qwater
mice * cice * ΔTice = mwater * cwater * ΔTwater
Substituting the given values:
150g * 2.09 J/g°C * (Tf - 0°C) = 300g * 4.18 J/g°C * (50°C - Tf)
Now, we can solve for Tf:
(150 * 2.09 * Tf) = (300 * 4.18 * (50 - Tf))
314.1Tf = 627 * (50 - Tf)
314.1Tf = 31350 - 627Tf
941.1Tf = 31350
Tf = 31350 / 941.1
Tf ≈ 33.3°C
Therefore, the temperature of the mixture is approximately 33.3°C.