Some very hot rocks have a temperature of #370 ^o C# and a specific heat of #90 J/(Kg*K)#. The rocks are bathed in #108 L# of boiling water. If the heat of the rocks completely vaporizes the water, what is the minimum combined mass of the rocks?

The heat transferred from the hot rocks to the water, is equal to the heat used to evaporate the water.

Heat of vaporisation of water

The specific heat of the rocks is

The mass of the rocks is

To find the minimum combined mass of the rocks, we can use the formula:

Q = mcΔT

Where:
Q = heat transferred (in joules)
m = mass (in kilograms)
c = specific heat capacity (in J/(kg*K))
ΔT = change in temperature (in Kelvin)

First, we need to calculate the heat transferred from the rocks to the water. Since the water is being completely vaporized, we'll use the heat of vaporization formula:

Q = mL

Where:
Q = heat transferred (in joules)
m = mass of water (in kilograms)
L = heat of vaporization (in joules per kilogram)

The heat of vaporization of water is approximately 2260 kJ/kg.

Now, we can set up the equation:

Q (from rocks) = Q (to vaporize water)

mcΔT = mL

Rearranging the equation to solve for mass (m) of the rocks:

m = mL / (cΔT)

Substituting the given values:
m = (2260 kJ/kg * 108 L) / (90 J/(kg*K) * 370 K)

Converting liters to kilograms:
1 liter of water = 1 kilogram
108 liters of water = 108 kilograms

Now, calculate the mass:
m = (2260 kJ/kg * 108 kg) / (90 J/(kg*K) * 370 K)

m ≈ 70.49 kg

So, the minimum combined mass of the rocks is approximately 70.49 kilograms.