Answer to Question #191741 in Chemistry for Jade

Question

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?

Answer 1

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:

-q_{lost by glass} = q_{gained by water}

 

q = m × C × ΔT

q = m × C × (T_f - T_i),

where:

q = amount of heat energy gained or lost by substance

m = mass of sample

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

T_f = final temperature

T_i = initial temperature

 

2) Therefore:

q_{lost by glass} = (45.7 g) × (C_glass) × (24.2 - 100.0)°C

q_{lost by glass} = -3464.06 × C_glass

 

q_{gained by waterÂ} = (250.0 g) × (4.20 J ^oC^-1 g^-1) × (24.2 - 22.5)°C

q_{gained by waterÂ} = 1785

 

-q_{lost by glass} = q_{gained by water}

-(-3464.06 × C_glass) = 1785

C_glass = 1785 / 3464.06 = 0.5153

C_glass = 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.