Question

23.a) A student performed a calorimetry experiment. He combined 50.0 mL of water at 82.2 ̊C and 50.0 mL of water at 34.4 ̊C in a coffee cup calorimeter. The final temperature of the water was 56.5 ̊C. What was the heat capacity of the calorimeter?

b) The same calorimeter was used to find the specific heat of a metal. He heated 96.4 grams of the metal in a boiling water bath (98.7 ̊C). After adding this hot metal to 100.0 mL of 30.4 ̊C water in the calorimeter, the final temperature reached 32.0 ̊C. What is the metal’s identity? Specific heat values are given in the reference page.

REFERENCE: specific heat of water = 4.184 J/(g∙ ̊C) specific gravity of water = 1.00 specific heat of water = 4.184 J/(g∙ ̊C) specific heat of Au = 0.129 J/(g∙ ̊C) specific heat of Cu = 0.385 J/(g∙ ̊C) specific heat of Fe = 0.450 J/(g∙ ̊C) specific heat of steam = 2.03 J/(g∙ ̊C) specific heat of ice = 2.09 J/(g∙ ̊C)

Answer 1

A) Heat capaity of calorimeter

Qgain = - Qlost

Qgain = cold water + calorimeter = mcold * Cwater * (Tf - Tcold) + Ccal * (Tf- Tcal)

Qlost = hot water =  mhot* Cwater * (Tf - Thot)

substitute in equation

mcold * Cwater * (Tf - Tcold) + Ccal * (Tf- Tcal)=  mhot* Cwater * (Tf - Thot)

substitute data, assume D = 1 g/mL

50* 4.184* (56.5- 34.4) + Ccal* (56.5- 34.4)= -50* 4.184* (56.5- 82.2)

Solv efor CCal

4623.32 + 22.1 * Ccal = 5376.44

Ccal = (5376.44 - 4623.32 )/22.1 = 34.0778 J/C

B)

m = 96.4 g of metal

Tmetal = 98.7

m water = 100 g

Twater = 30.4

Tf = 32

Find the metal Speicfi cheat

-Qmetal = Qwater

-mmetal * Cmetal * (Tf-Tmetal) = mwater * Cwater * (Tf-Twater)

-96.4* Cmetal * (32-98.7) = 100* 4.184* (32-30.4)

Cmetal =  100* 4.184* (32-30.4) / (-96.4* (32-98.7))

Cmetal = 0.1041139 J/gCnearest answer from the list is Gold, this must be Au, with 0.129 J/gC specific heat