A 10 g sample of water at 24 Celsius is mixed with 30.0 g of water at 48 Celsius. Calculate the final temperature of the mixture assuming no heat loss to the surroundings. The specific heat of water is 4.18 j/g•C
heat lost by water + heat gained by cool water = 0
[mass warm H2O x specific heat H2O x (Tfinal - Tinitial)] + [mass cool H2O x specific heat H2O x (Tfinal-Tinitial)] = 0. Tfinal is the only unknown. Substitute and solve for it.
28.0
To calculate the final temperature of the mixture, we can use the principle of conservation of energy. The total energy gained by the cooler water will be equal to the total energy lost by the warmer water.
First, let's determine the energy gained by the cooler water. We can use the formula: Q = m * c * ΔT, where Q is the energy gained or lost, m is the mass in grams, c is the specific heat capacity, and ΔT is the change in temperature.
For the cooler water:
m1 = 10 g (mass of cooler water)
c = 4.18 J/g•C (specific heat of water)
ΔT1 = Tf - 24 (change in temperature)
Now, let's determine the energy lost by the warmer water:
m2 = 30.0 g (mass of warmer water)
c = 4.18 J/g•C (specific heat of water)
ΔT2 = 48 - Tf (change in temperature)
Since the energy gained by the cooler water equals the energy lost by the warmer water, we have:
m1 * c * ΔT1 = m2 * c * ΔT2
Substituting the values, we can rearrange the equation to solve for Tf:
10 * 4.18 * (Tf - 24) = 30.0 * 4.18 * (48 - Tf)
Now, simplify and solve for Tf:
41.8 * Tf - 1003.2 = 125.4 * Tf - 1507.2
-83.6 = 83.6 * Tf
Tf = -1 degree Celsius
So the final temperature of the mixture will be -1 degree Celsius.