Question

Tap water has a temperature of 60°C. The same cup of tap water is used for

3 other experiments before Dr. Woods finally gets it. Each experiment raises
the temperature of the water 12.5°C without allowing it to cool between
experiments. In his test, Dr. Woods has to lower the water 98°C before it
freezes. Find the freezing point of the tap water and compare it to the normal
freezing point of water.
He then raises the temperature of the same water 99.25°C, which causes it
to boil. Compare its boiling point to the normal boiling point of water.

Answer 1

To solve this problem, we can use the formula:

Q = mcΔT

where Q is the heat energy transferred, m is the mass of the water, c is the specific heat capacity of water (4.184 J/g°C), and ΔT is the change in temperature.

For the first part of the problem, we can use the formula to find the freezing point of the tap water:

Q1 = mcΔT1

where ΔT1 = 98°C - 60°C = 38°C

We assume that the mass of the water is the same in all experiments and in Dr. Woods' test, so we can use the same value of m throughout.

Q1 = mcΔT1 = m(4.184 J/g°C)(38°C)

Q1 = 1593.792 J

Each of the 3 other experiments raised the temperature of the water by 12.5°C, so the total change in temperature is:

ΔT2 = 12.5°C x 3 = 37.5°C

The heat energy transferred in these experiments is:

Q2 = mcΔT2 = m(4.184 J/g°C)(37.5°C)

Q2 = 1566.9 J

The total heat energy transferred before the water freezes is:

Q = Q1 + Q2 = 1593.792 J + 1566.9 J = 3160.692 J

We can now use the formula to find the freezing point of the tap water:

Q = mcΔTf

where ΔTf = T - 0°C

Assuming the same mass of water as before, we get:

3160.692 J = m(4.184 J/g°C)(60°C - 0°C)

m = 18.995 g

ΔTf = T - 0°C = 3160.692 J / (18.995 g x 4.184 J/g°C) ≈ 40°C

So, the freezing point of the tap water is approximately 40°C, which is 8 degrees higher than the normal freezing point of water (0°C).

For the second part of the problem, we can use the same formula to find the boiling point of the tap water:

Q = mcΔTb

where ΔTb = 100°C - T

Assuming the same mass of water as before, we get:

Q = mcΔTb = m(4.184 J/g°C)(99.25°C)

Q = 4158.83 J

4158.83 J = m(4.184 J/g°C)(T - 0°C)

T = 100°C + 4158.83 J / (18.995 g x 4.184 J/g°C) ≈ 116°C

So, the boiling point of the tap water is approximately 116°C, which is 16 degrees higher than the normal boiling point of water (100°C).