# Answer to Question #191741 in Chemistry for Jade

Question #191741

2) A 45.7 g sample of glass was brought to thermal equilibrium with boiling water and then transferred to 250.0 g of water that was at 22.5 °C. This combination reached thermal equilibrium at 24.2 °C. What is the specific heat capacity of glass?

1
Expert's answer
2021-05-11T05:57:22-0400

Water boils at 100°C

The specific heat capacity of water is 4.20 J oC-1 g-1.

Solution:

1) This problem can be summarized thusly:

-qlost by glass = qgained by water

q = m × C × ΔT

q = m × C × (Tf - Ti),

where:

q = amount of heat energy gained or lost by substance

m = mass of sample

C = specific heat capacity (J oC-1 g-1)

Tf = final temperature

Ti = initial temperature

2) Therefore:

qlost by glass = (45.7 g) × (Cglass) × (24.2 - 100.0)°C

qlost by glass = -3464.06 × Cglass

qgained by water = (250.0 g) × (4.20 J oC-1 g-1) × (24.2 - 22.5)°C

qgained by water = 1785

-qlost by glass = qgained by water

-(-3464.06 × Cglass) = 1785

Cglass = 1785 / 3464.06 = 0.5153

Cglass = 0.5153 J oC-1 g-1 = 515.3 J oC-1 kg-1

Answer: The specific heat capacity of glass is 515.3 J oC-1 kg-1.

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