A hot water tap of a bath delivers water at 85⁰C at a rate of 8Kg per minute. The cold water tap at the bath delivers water at 30⁰C at the rate of 20kg per minutes. If both tapes are at a time of 5 minutes. Calculate the final temperature o f the bath water, ignoring heat loses to the surrounding in kelvin's

Answer 1

To find the final temperature of the bath water, we have to apply the principle of conservation of energy. The total heat gained by the water in the bath must be equal to the total heat lost by the hot water tap and gained by the cold water tap.

Let:
m1 = mass of hot water tap = 8 kg/min
m2 = mass of cold water tap = 20 kg/min
T1 = initial temperature of hot water = 85⁰C
T2 = initial temperature of cold water = 30⁰C
Tf = final temperature of bath water

The total heat gained or lost by an object can be calculated using the equation:

Q = mcΔT

Where:
Q = heat transfer
m = mass
c = specific heat capacity
ΔT = change in temperature

Assuming the specific heat capacity of water is 4200 J/kg°C, the heat transfer for the hot water tap is:
Q1 = m1 * c * (Tf - T1)

And for the cold water tap is:
Q2 = m2 * c * (Tf - T2)

Since there is no heat transfer to the surroundings, the heat lost by the hot water tap should be equal to the heat gained by the cold water tap, i.e., Q1 = Q2.

Now, we can set the equations equal to each other:
m1 * c * (Tf - T1) = m2 * c * (Tf - T2)

Substitute the known values:
8 * 4200 * (Tf - 85) = 20 * 4200 * (Tf - 30)

Simplify and solve for Tf:
33600Tf - 2856000 = 84000Tf - 1260000
-50400Tf = -1416000
Tf = 28.1⁰C

Therefore, the final temperature of the bath water after 5 minutes will be approximately 28.1⁰C.