Water with a mass of 0.500 kg at 15.0 degree C is mixed with 0.800 kg of water at 35.0 degree C. What is the final temperature T_f of the mixture?
27.3 c
the sum of heats gained is zero.
.5*c*(Tf-15)+.8*c*(Tf-35)=0
solve for Tf
in a laboratory a student mixes 300 g water at 80 degree with 400 g water at 20 degree
Well, it seems we have a hot date between two waters! Let's see if they can find their temperature match.
To find the final temperature, we can use the principle of conservation of energy. The total energy before mixing is equal to the total energy after mixing.
The energy of the first water is given by its mass (0.500 kg) multiplied by its specific heat capacity (4,184 J/kg°C) and the change in temperature (T_f - 15.0 °C).
Similarly, the energy of the second water is given by its mass (0.800 kg) multiplied by its specific heat capacity (4,184 J/kg°C) and the change in temperature (T_f - 35.0 °C).
Since the total energy before mixing is equal to the total energy after mixing, we can set up the equation:
(mass1 * specific heat1 * (T_f - 15.0)) + (mass2 * specific heat2 * (T_f - 35.0)) = 0
Now let's solve this equation together to find the final temperature T_f. Who says science can't be fun?
To find the final temperature of the mixture, we can use the principle of conservation of energy.
The equation we will use is:
(m1 * c1 * ΔT1) + (m2 * c2 * ΔT2) = 0
where:
- m1 and m2 are the masses of water in kg
- c1 and c2 are the specific heat capacities of water in J/kg·K
- ΔT1 and ΔT2 are the changes in temperature in K
We are given:
- m1 = 0.500 kg
- c1 = 4186 J/kg·K (specific heat capacity of water)
- ΔT1 = (Tf - 15.0 °C) (change in temperature of the first batch of water)
- m2 = 0.800 kg
- c2 = 4186 J/kg·K (specific heat capacity of water)
- ΔT2 = (Tf - 35.0 °C) (change in temperature of the second batch of water)
Now, let's substitute the values into the equation and solve for Tf:
(0.500 kg * 4186 J/kg·K * (Tf - 15.0 °C)) + (0.800 kg * 4186 J/kg·K * (Tf - 35.0 °C)) = 0
Simplifying the equation:
2093(Tf - 15.0) + 3349.6(Tf - 35.0) = 0
2093Tf - 31395 + 3349.6Tf - 117246 = 0
5442.6Tf - 148641 = 0
5442.6Tf = 148641
Tf = 148641 / 5442.6
Tf ≈ 27.3 °C
Therefore, the final temperature T_f of the mixture is approximately 27.3 °C.