Question

A 50.8g sample of glass, which has a specific heat capacity of 0.670·J·g?1°C?1, is put into a calorimeter (see sketch at right) that contains 150.0g of water. The temperature of the water starts off at 22.0°C. When the temperature of the water stops changing it's 25.6°C. The pressure remains constant at 1atm. Calculate the initial temperature of the glass sample. Be sure your answer is rounded to the correct number of significant digits.

Answer 1

Step-by-step explanation:

mass of glass m = 50.8 g

specific heat s = .67 J /g °C

initial temperature = t

Loss of heat by glass

= m s Δ t

= 50.8 x .67 x ( t - 25.6 )

= 34.036 x ( t - 25.6 )

Gain of heat by water

= 150 x 4.2 x ( 25.6 - 22.0 )

= 2268

Heat gain = heat lost

2268 = 34.036 x ( t - 25.6 )

2268 = 34.036 t - 871.32

34.036 t = 3139.32

t = 92.23°C .