Question

In the laboratory a "coffee cup" calorimeter, or constant pressure calorimeter, is frequently used to determine the specific heat of a solid, or to measure the energy of a solution phase reaction. A student heats 62.08 grams of magnesium to 97.96 °C and then drops it into a cup containing 77.81 grams of water at 23.19 °C. She measures the final temperature to be 35.60 °C. The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.79 J/°C. Assuming that no heat is lost to the surroundings calculate the specific heat of magnesium.

Answer 1

The specific heat of magnesium is 1.04 J/g°C.

Step-by-step explanation:

Heat lost by the magnesium = Q

Mass of the magnesium = m = 62.08 g

Heat capacity of magnesium= c = ?

Initial temperature of the magnesium =
`T_1=97.96^oC`

Final temperature of the magnesium= T = 35.60 °C

`Q=mc* (T-T_1)`

Heat absorbed by coffee cup calorimeter = Q'

Heat capacity of coffee cup calorimeter = C = 1.79 J/°C

Initial temperature of coffee cup calorimete =
`T_2` = 23.19°C

Final temperature of coffee cup calorimete = T = 35.60 °C

`Q'=C* (T-T_2)`

Heat absorbed by the water = q

Mass of water = m' = 77.81 g

Heat capacity of water = c' = 4.18 J/g°C

Initial temperature of water =
`T_2` = 0°C

Final temperature of water = T

`q=m'* c'* (T_2-T)`

According law of conservation of energy , energy lost by coffee will equal to heat required to raise temperature of water and coffee cup calorimeter.

`-Q=Q'+q`

`-(mc* (T-T_1))=C* (T-T_2)+m'* c'* (T-T_2)`

`62.08 g* c(97.96^oC-35.60^oC)=1.79 J/^oC* (35.60^oC-23.19^oC)+77.81g* 4.18 J/g^oC* (35.60^oC-23.19^oC)`

On solving we get:

c = 1.04 J/g°C

The specific heat of magnesium is 1.04 J/g°C.