Answer to Question #87158 in Molecular Physics | Thermodynamics for MUHAMMED

Question

A piece of copper of mass 300g at a temperature of 950degree celcius is quickly transferred to a vessel of negligible thermal capacity containing 250g of water at 25degree celcius. if the final steady temperature of the mixture is 100degree celcius. calculate the mass of water that will boil away

Answer 1

According to the heat balance equation,

$$Q = Q_1 +Q_2,$$

where Q is the amout of heat released by copper due to cooling, Q₁ is the amount of heat that was spent on water heating (up to 100 degree Celsius), Q₂ is the amount of heat spent on partial water vaporization. Their explicit expressions can be written as follows:

$$Q=cm \Delta T, \\ Q_1 = c_1 m_1 \Delta T_1,\\ Q_2 = L \Delta m,$$

where c and c₁ are the specific heats of copper and water correspondingly, m and m₁ - their masses, (delta m) is the amount of vaporized water, L is the specific heat of vaporization of water.

As a result, we derive:

$$\Delta m = \frac{c m \Delta T - c_1 m_1 \Delta T_1}{L}$$

Substituting the numerical values, we obtain:

$$\Delta m = \frac{385 \cdot 0.3 \cdot (950-100) - 4186 \cdot 0.25 \cdot (100-25)}{2258 \cdot 10^3} \approx 8.7 \cdot 10^{-3} \, kg \approx 9 \, g$$

Answer: 9 g.