A 200 mL of water at 30°C is mixed with 300 mL of water at

Question

20°C in a perfectly insulated container. Assuming uniform
mixing, determine the final temperature of the system. Assume
the density equal to 1000 kg/m3
.

Answer 1

To solve this problem, we can use the principle of conservation of energy.

The heat gained by the cooler water (300 mL at 20°C) will be equal to the heat lost by the hotter water (200 mL at 30°C). We can express this using the equation:

Q1 = Q2

Where,
Q1 is the heat gained by the cooler water
Q2 is the heat lost by the hotter water

To calculate the heat gained or lost, we can use the formula:

Q = mcΔT

Where,
Q is the heat gained or lost in joules
m is the mass of the water in kg
c is the specific heat capacity of water (approximately 4200 J/kg°C)
ΔT is the change in temperature in °C

For the cooler water (300 mL at 20°C), the mass (m1) can be calculated using the density of water:

m1 = (density of water) * (volume of water)
= 1000 kg/m^3 * 0.3 kg
= 300 kg

ΔT1 = final temperature - initial temperature
= final temperature - 20°C

Substituting these values into the equation for heat gained by the cooler water (Q1), we get:

Q1 = (300 kg) * (4200 J/kg°C) * ΔT1

Similarly, for the hotter water (200 mL at 30°C), the mass (m2) can be calculated:

m2 = (density of water) * (volume of water)
= 1000 kg/m^3 * 0.2 kg
= 200 kg

ΔT2 = initial temperature - final temperature
= 30°C - final temperature

Substituting these values into the equation for heat lost by the hotter water (Q2), we get:

Q2 = (200 kg) * (4200 J/kg°C) * ΔT2

Since Q1 = Q2, we can equate the two equations and solve for the final temperature:

(300 kg) * (4200 J/kg°C) * ΔT1 = (200 kg) * (4200 J/kg°C) * ΔT2

Dividing both sides by (4200 J/kg°C) and canceling out the kg term, we get:

(300) * ΔT1 = (200) * ΔT2

Dividing both sides by 300 and rearranging the terms, we get:

ΔT1 = (2/3) * ΔT2

Substituting the expression for ΔT1 into the equation, we get:

(2/3) * ΔT2 = ΔT2 - 20

Multiplying both sides by 3 to eliminate the fraction, we get:

2 * ΔT2 = 3 * ΔT2 - 60

Simplifying the equation, we get:

ΔT2 = 60

Substituting this value back into the expression for ΔT1, we get:

ΔT1 = (2/3) * 60
= 40

Finally, we can calculate the final temperature:

Final temperature = initial temperature - ΔT1
= 30°C - 40°C
= -10°C

Therefore, the final temperature of the system is -10°C.