A 30 g sample of water at 280 K is mixed with 50 g water at 330 K. Calculate the final temperature of the mixture assuming no heat loss to the surroundings.
let C = specific heat of water
heat gained = C*30*(T-280)
heat lost = C * 50(330-T)
so
30(T-280) = 50 (330-T)
To calculate the final temperature of the mixture, we can use the principle of conservation of energy. The total heat gained by the cold water (30 g at 280 K) will be equal to the total heat lost by the hot water (50 g at 330 K). This can be expressed using the equation:
m1 * c1 * (Tf - T1) = m2 * c2 * (T2 - Tf)
Where:
m1 = mass of the cold water (30 g)
c1 = specific heat capacity of water
Tf = final temperature of the mixture (to be determined)
T1 = initial temperature of the cold water (280 K)
m2 = mass of the hot water (50 g)
c2 = specific heat capacity of water
T2 = initial temperature of the hot water (330 K)
First, we need to find the specific heat capacity of water. The specific heat capacity of water is approximately 4.18 J/g°C. We can substitute this value into our equation:
30 * 4.18 * (Tf - 280) = 50 * 4.18 * (330 - Tf)
Now, we can solve this equation to find the final temperature (Tf) of the mixture:
125.4 * Tf - 35712 = 2089 * Tf - 67839
125.4 * Tf - 2089 * Tf = -67839 + 35712
-1963.6 * Tf = -32127
Tf = -32127 / -1963.6
Tf = 16.37°C
Therefore, the final temperature of the mixture, assuming no heat loss to the surroundings, is approximately 16.37°C.