A person tries to heat up her bath water by adding 5.0 L of water at 80 degrees celcious to 60 L of water at 30 degrees celcious. What is the final temperature of the water?

Show work please !

To determine the final temperature of the water, we can use the concept of heat transfer, assuming that there is no heat loss to the surroundings.

The heat gained by the cold water (Q1) is equal to the heat lost by the hot water (Q2).

The formula to calculate heat transfer is:

Q = mcΔT

Where,
Q = heat transfer (in Joules),
m = mass (in kg),
c = specific heat capacity (in J/kg°C), and
ΔT = change in temperature (in °C).

Let's calculate the heat gained by the cold water (Q1):

Q1 = mcΔT
= (m1)(c)(ΔT1)
= (60 kg)(4.18 J/kg°C)(Tf - 30°C) --(equation 1)

Now let's calculate the heat lost by the hot water (Q2):

Q2 = mcΔT
= (m2)(c)(ΔT2)
= (5 kg)(4.18 J/kg°C)(80°C - Tf) --(equation 2)

According to the principle of conservation of energy, heat gained should be equal to heat lost:

Q1 = Q2

Now, we can substitute equations 1 and 2 into the conservation equation:

(60 kg)(4.18 J/kg°C)(Tf - 30°C) = (5 kg)(4.18 J/kg°C)(80°C - Tf)

Simplifying the equation:

(60)(Tf - 30) = (5)(80 - Tf)

Expanding:

60Tf - 1800 = 400 - 5Tf

Combining like terms:

65Tf = 2200

Dividing by 65:

Tf ≈ 33.8°C

Therefore, the final temperature of the water will be approximately 33.8°C.