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 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.

Answer 1

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

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