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.

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 =

Final temperature of the magnesium= T  = 35.60 °C

Heat absorbed by coffee cup calorimeter  = Q'

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

Initial temperature of coffee cup calorimete = = 23.19°C

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

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 = = 0°C

Final temperature of water = 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.

On solving we get:

c = 1.04 J/g°C

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

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm