Question

Indigenous people sometimes cook in watertight baskets by placing hot rocks into water to bring it to a boil. What mass of 500ºC rock must be placed in 4.00 kg of 15.0ºC water to bring its temperature to 100ºC, if 0.0250 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. (answer in kg)

Answer 1

Answer:

m = 4.65 kg

Explanation:

As we know that the mass of the water that evaporated out is given as

$m = 0.0250 kg$

so the energy released in form of vapor is given as

$Q = mL$

$Q = (0.0250)(2.25 \times 10^6)$

$Q = 56511 J$

now the heat required by remaining water to bring it from 15 degree to 100 degree

$Q_2 = ms\Delta T$

$Q_2 = (4 - 0.025)(4186)(100 - 15)$

$Q_2 = 1.41\times 10^6J$

total heat required for above conversion

$Q = 56511 + 1.41 \times 10^6 = 1.47 \times 10^6 J$

now by heat energy balance

heat given by granite = heat absorbed by water

$m(790)(500 - 100) = 1.47 \times 10^6$

$m = 4.65 kg$