Question

My hot water system maintains a volume of 200L of water, which it heats at a rate of 1 ◦C per minute, up to a maximum temperature of 60◦C. On a winter's morning, the tank is full of water at this maximum temperature. While the shower is on, the tank is relled with water from the cold water source, which is at 10◦C. Assume the water in the tank maintains a uniform temperature (because is well-mixed). In my shower, the taps x the rate (in L per minute) at which cold and hot water come from their respective sources.

(a) At the start of my shower, I set the taps so that the water comes out at 35◦C initially, at a rate of 20L per minute. Assuming I leave both taps at their initial rates, how long would it take before the water temperature drops to 25◦C?

(b) We have two showers in our household. If someone else started their shower at the same time, at the same initial temperature and ow rate, how long would it now take before the water temperature drops to 25◦C?

Answer 1

When two liquid gets mixed, of different temperature
Q lost = Q gain (heats )
Q = (mass)*(delta T)*Cp
Cp for water is = 4.2 J/g.K
(for cold water) delta T = 35 - 25 = 10
(for hot water) delta T = 60-25 = 35
now mass will be unknows in this case, and that can corroesponds to Volume per minute.

mass of
Q lost by hot water = Q gained by cold water
mass1*10*4.2 = mass2*35*4.2
mass1(hot) = 3.5 * mass2(cold)
proportion of liquids at end of temperature 25 is : 3.5
m1/m2 = 3.5 or same is said for volumes also
as Q = V(T hot - T final) _____________ _ _ _ _ heat gained, when volume is given

Now based on rate of water dispence from each water tank, we can calculate time to reach this ratio
V1/V2 = 3.5
V1 = 20 L/minute * time
V2 = 10 L

20t/10 = 3.5
time = 3.5*10 / 20 = 1.75 minutes

now in case one more shower is on, with same flow rate of water as first one.
The time to reach the temperature will be doubled as the amount of water dispensed is doubled.
so t = 3.5 minutes
Answer