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

Answer:

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

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

$Q=mc\times (T-T_1)$

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

$Q'=C\times (T-T_2)$

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

$q=m'\times c'\times (T_2-T)$

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

$-Q=Q'+q$

$-(mc\times (T-T_1))=C\times (T-T_2)+m'\times c'\times (T-T_2)$

$62.08 g\times c(97.96^oC-35.60^oC)=1.79 J/^oC\times (35.60^oC-23.19^oC)+77.81g\times 4.18 J/g^oC\times (35.60^oC-23.19^oC)$

On solving we get:

c = 1.04 J/g°C

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