Question

Indigenous people sometimes cooked in watertight baskets by placing enough hot rocks into the water to bring it to a boil. What mass of 500°C rock must be placed in 4.24 kg of 29°C water to bring its temperature to 100°C, if 0.0440 kg of water escapes as vapor from the initial sizzle?

You may neglect the effects of the surroundings and take the average specific heat of the rocks to be that of granite, which is 0.20 kcal/(kg * °C).
The specific heat of water is 1.00 kcal/(kg * °C) and the latent heat of vaporization is 539 kcal/kg.

Answer 1

The rock has a mass of 4.02 kg

Step-by-step explanation:

Step 1: Data given

Mass of the rock = TO BE DETERMINED

Temperature of the rock = 500 °C

Mass of the water =4.24 kg

⇒ loses 0.044kg as vapor

Initial temperature of the water = 29°C

Final temperature = 100°C

Specific heat of rock = 0.20 kcal/kg °C

Specific heat of water = 1kcal/kg°C

Latent heat of vaporization = 539 kcal/kg

Step 2: formules

Qlost,rock + Qgained,water = 0

Qtotal,water = Qwater +Qvapor

Step 3: Calculate Qvapor

Qvapor = mass of vapor * Latent heat of vapor

Qvapor = 0.044kg * 539 kcal/kg = 23.716 kcal

Step 4: Calculate Qwater

Qwater = mass of water * specific heat * Δtemperature

Qwater = 4.196 kg * 1kcal/kg°C *( 100-29)

Qwater = 297.916 kcal

Step 5: Calculate Qwater,total

Qwater,total = Qwater + Qvapor

Qwater,total = 23.716 kcal + 297.916 = 321.632 kcal

Step 6: Calculate Qrock

Qrock = mass of rock * specific heat rock * Δtemperature

Qrock = mass of rock * 0.20 kcal/kg°C * (100-500)

Qrock = mass of rock * -80 kcal/kg

Step 7: Calculate mass of rock

Qlost,rock + Qgained,water = 0

Qlost,rock = -Qgained,water

mass of rock * -80 kcal/kg = -321.632 kcal

mass of rock = 4.02 kg

The rock has a mass of 4.02 kg