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
A student heats 60.93 grams of gold to 98.87 °C and then drops it into a cup containing 79.68 grams of water at 24.46 °C. She measures the final temperature to be 26.11 °C.
The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.54 J/°C.
Assuming that no heat is lost to the surroundings calculate the specific heat of gold.
Specific Heat (Au) = _______J/g°C.
Let us denote H2O by symbol 1, gold by symbol 2 and calorimeter by symbol 3
m1 = 79.68 g
T1 = 24.46 oC
C1 = 4.18 J/goC
m2 = 60.93 g
T2 = 98.87 oC
C2 = to be calculated
T = 26.11 J/goC
C3 = 1.54 J/g
we have below equation to be used:
heat lost by 2 = heat gained by 1 and 3
m2*C2*(T2-T) = m1*C1*(T-T1) + C3*(T-T1)
60.93*C2*(98.87-26.11) = 79.68*4.18*(26.11-24.46) + 1.54*(26.11-24.46)
4433.2668*C2 = 549.553 + 2.541
C2= 0.1245 J/goC
Answer: 0.125 J/goC
Feel free to comment below if you have any doubts or if this answer do not work
Get Answers For Free
Most questions answered within 1 hours.