Question

A 52.4 g sample of brass is put into a calorimeter (see sketch at right) that contains 150.0 g of water. The brass sample starts off at 95.1 °C and the temperature of the water starts off at 15.0 °C. When the temperature of the water stops changing it's 17.6°C. The pressure remains constant at 1 atm. container water Calculate the specific heat capacity of brass according to this experiment.

Answer 1

Answer:

The specific heat capacity of brass is 0.402 J/g°C

Explanation:

Step 1: Data given

Mass of the sample of brass = 52.4 grams

Mass of water = 150.0 grams

Initial temperature of brass sample = 95.1 °C

Initial temperature of water = 15.0 °C

Final temperature = 17.6°C

Pressure = 1atm

Specific heat capacity = 4.184 J/g°C

Step 2: Calculate specific heat capacity of brass sample

Heat lost by the brass sample = heat won by water

Q = m*c*ΔT

Qbrass = -Qwater

m(brass)*c(brass)*ΔT(brass) = -m(water) *c(water) * ΔT(water)

⇒mass of brass sample = 52.4 grams

⇒ c(brass) = TO BE DETERMINED

⇒ ΔT = T2 - T1 = 17.6 °C - 95.1 °C = -77.5 °C

⇒ mass of water = 150.0 grams

⇒ c(water ) = 4.184 J/g°C

⇒ ΔT(water) = 17.6 °C - 15 °C = 2.6°C

52.4g * c(brass) * (-77.5°C) = -150.0g * 4.184 J/g°C * 2.6 °C

-4061 * c(brass) = -1631,76

c(brass) = 0.402 J/g°C

The specific heat capacity of brass is 0.402 J/g°C