Mary mixes 5.00 lb of water at 200°F with 7.00 lb of water at 65.0°F. Find the final temperature of the mixture.

-Specific Heat of water is 1.00 BTU/lbs°F

To find the final temperature of the mixture, we can use the principle of conservation of energy. The total heat gained by the cold water will be equal to the total heat lost by the hot water.

The equation to calculate heat (Q) is as follows:

Q = mcΔT

Where:
- Q is the heat energy
- m is the mass of the substance
- c is the specific heat
- ΔT is the change in temperature

First, let's calculate the heat gained by the cold water (Q_cold):

Q_cold = mcΔT

m = 7.00 lb (mass of cold water)
c = 1.00 BTU/lb°F (specific heat of water)
ΔT = final temperature - initial temperature = final temperature - 65.0°F

Q_cold = 7.00 lb * 1.00 BTU/lb°F * (final temperature - 65.0°F)

Next, let's calculate the heat lost by the hot water (Q_hot):

Q_hot = mcΔT

m = 5.00 lb (mass of hot water)
c = 1.00 BTU/lb°F (specific heat of water)
ΔT = 200.0°F - final temperature

Q_hot = 5.00 lb * 1.00 BTU/lb°F * (200.0°F - final temperature)

Since the total heat gained by the cold water is equal to the total heat lost by the hot water, we can set up an equation:

Q_cold = Q_hot

7.00 lb * 1.00 BTU/lb°F * (final temperature - 65.0°F) = 5.00 lb * 1.00 BTU/lb°F * (200.0°F - final temperature)

Simplifying the equation, we get:

7.00 (final temperature - 65.0) = 5.00 (200.0 - final temperature)

Now we can solve for the final temperature:

7.00 * final temperature - 455.0 = 1000.0 - 5.00 * final temperature

12.00 * final temperature = 1000.0 + 455.0

12.00 * final temperature = 1455.0

final temperature = 1455.0 / 12.00

final temperature ≈ 121.25°F

Therefore, the final temperature of the mixture is approximately 121.25°F.