Answer to Question #83470 in General Chemistry for Michael Alegado

Question

What is the mass of water (in g) at 100 °C that can be completely boiled into liquid water at 100°C by a 86 g aluminum block at temperature 306 °C? Assume the aluminum is capable of boiling the water until its temperature drops below 100 °C.

The heat capacity of aluminum is 0.903 J g-1 °C-1 and the heat of vaporization of water at 100°C is 40.7 kJ mol-1.

Answer 1

Answer on question #83470, Chemistry/ General Chemistry

What is the mass of water (in g) at 100 °C that can be completely boiled into liquid water at 100°C by a 86 g aluminum block at temperature 306 °C? Assume the aluminum is capable of boiling the water until its temperature drops below 100 °C.

The heat capacity of aluminum is 0.903 J g-1 °C-1 and the heat of vaporization of water at 100°C is 40.7 kJ mol-1.

Solution

1. Find the heat that is released when 86 g of aluminum block is cooled from 306C to 100C

Q= cm(T₂ - T₁)

Q_released = 0.903 J/(g*C) *86g*(100C - 306C) = -15998 J = -16.0 kJ

2. Find mass of water:

The heat released by aluminum is absorbed by water

Q_released = Q_absorbed

Q_absorbed= ΔH_vap*n

Q_absorbed = 40.7 kJ/mol * n

Then 16.0 kJ = 40.7 kJ/mol * n

n= 0.393 mol

m= M*n

M(H₂O) = 1.01*2 +16.00 = 18.02 g/mol

m(H₂O) = 18.02 g/mol *0.393 mol = 7.08 g

Answer: 7.08 g