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

Question

Quizplus · Accepted Answer

First, we need to calculate the final temperature of the mixture, which can be done using the formula:

m1c1(T1 - T2) = m2c2(T2 - T1)

Where m1 is the mass of the metal, c1 is its specific heat, T1 is its initial temperature, T2 is the final temperature, m2 is the mass of water, and c2 is its specific heat.

Substituting the values:

3 kg x 0.1 kcal/kg°C x (600°C - T2) = 1 kg x 1 kcal/kg°C x (T2 - 20°C)

1800 - 0.3T2 = T2 - 20

1.3T2 = 1820

T2 = 140°C

So the equilibrium temperature of the mixture is 140°C, which means that some of the water will turn into steam.

To calculate how much water is turned into steam at 100°C, we use the formula:

m = (Q / hfg)

Where m is the mass of water turned into steam, Q is the heat absorbed by the water, and hfg is the latent heat of vaporization of water, which is 540 kcal/kg.

The heat absorbed by the water can be calculated using:

Q = m2c2(T2 - 100°C)

Substituting the values:

Q = 1 kg x 1 kcal/kg°C x (140°C - 100°C) = 40 kcal

Finally, substituting in the formula for m:

m = (40 / 540) = 0.074 kg = 74 g

Therefore, the answer is C) 100°C and 130 g of steam.

A 3-Kg Mass of Metal of Specific Heat = 0