A 3-kg mass of metal of specific heat = 0.1 kcal/kg $\circ$ C at a temperature of 600 $\circ$ C is dropped into 1.0 kg water at 20 $\circ$ C.With no heat losses to the surroundings,determine the equilibrium temperature of the mixture,and if it is 100 $\circ$ C,calculate what mass of water is turned into steam at this temperature.

Question

Quizplus · Accepted Answer

First, we need to calculate the heat lost by the metal and the heat gained by the water:
Q lost = m metal × c metal × (T initial - T final)
Q lost = 3 kg × 0.1 kcal/kg°C × (600°C - T final)
Q gained = m water × c water × (T final - T initial)
Q gained = 1 kg × 1 kcal/kg°C × (T final - 20°C)

Since there is no heat loss to the surroundings, Q lost = Q gained:
3 kg × 0.1 kcal/kg°C × (600°C - T final) = 1 kg × 1 kcal/kg°C × (T final - 20°C)
180 - 0.3 T final = T final - 20
1.3 T final = 200
T final = 153.8°C or 308.8°F

Now, we need to use the heat gained by the water to calculate the mass of water turned into steam:
Q = m water × L vaporization
Q = 1 kg × 540 kcal/kg
Q = 540 kcal

The heat gained by the water is:
Q gained = m water × c water × (T final - 20°C)
Q gained = m water × 1 kcal/kg°C × (153.8°C - 20°C)
Q gained = m water × 133.8 kcal

Since Q gained is greater than Q, some of the water turns into steam. The amount of water turned into steam is:
m water = Q / (c water × (T final - 20°C) + L vaporization)
m water = 540 / (1 kcal/kg°C × (153.8°C - 20°C) + 540 kcal/kg)
m water = 130 g

Therefore, the equilibrium temperature is 100°F and 130 g of water is turned into steam, so the best choice is C.

A 3-Kg Mass of Metal of Specific Heat = 0 ∘\circ∘

A 3-Kg Mass of Metal of Specific Heat = 0 $\circ$