Suppose you pour 0.310 kg of 20.0°C water into a 0.600 kg aluminum pan off the stove with a temperature of 167°C. Assume that the pan is placed on an insulated pad. What would be the final temperature (in °C) of the pan and water if 0.0300 kg of the water evaporated immediately, leaving the remainder to come to a common temperature with the pan?

Cw= 4186, Cal=900, Lvapor=2256000
Find final temperature

final temp = T

heat into water :
first .03 kg is brought to 100 and boiled off
heat into that water = .03 (2256000) + .03(4186)(100 - 20) = 77726
the rest of the water is .31 - .03 = .27 kg
so heat into it is
.27 (4186)(T-20) = 1130 (T-20)
so
total heat into water = 77726 +1130(T-20)

heat into pan = .6 (900)(167-T)
so
77726 +1130(T-20) = .6 (900)(167-T)

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

First, let's calculate the energy gained by the cold water:

Energy gained by water = mass of water remaining * specific heat capacity of water * change in temperature

mass of water remaining = initial mass of water - mass of water evaporated
= 0.310 kg - 0.0300 kg
= 0.28 kg

Change in temperature = final temperature - initial temperature

Now, let's calculate the energy lost by the hot pan:

Energy lost by pan = mass of pan * specific heat capacity of aluminum * change in temperature

mass of pan = 0.600 kg
specific heat capacity of aluminum (Cal) = 900
change in temperature = final temperature - initial temperature

Since the pan is on an insulated pad, there is no loss of heat to the surroundings. Therefore, the energy lost by the pan equals the energy gained by the water:

Energy lost by pan = Energy gained by water

Now we can equate the two expressions and solve for the final temperature:

mass of pan * specific heat capacity of aluminum * (final temperature - initial temperature) = mass of water remaining * specific heat capacity of water * (final temperature - initial temperature)

0.600 kg * 900 * (final temperature - 167°C) = 0.28 kg * 4186 * (final temperature - 20.0°C)

Now, expand and solve for the final temperature:

540 * (final temperature - 167°C) = 1170 * (final temperature - 20.0°C)

540 * final temperature - 90480 = 1170 * final temperature - 23400

630 * final temperature = 67080

final temperature = 67080 / 630

final temperature ≈ 106.67°C

Therefore, the final temperature of the pan and water mixture would be approximately 106.67°C.