Scoring Scheme: 3-3-2-1 Part I. For each trial, enter the amount of heat gained by the cool water, qcool water. The specific heat of water is 4.184 J/goC. Report your answer to 4 digits. Note: You should always carry 1 or 2 extra digits beyond the number of significant figures until your final calculation.

Hints: The specific heat of water is 4.184 J/g°C. Be careful of your algebraic sign here and remember that the change in temperature is equal to the final temperature minus the initial temperature.

qrxn = - (qwater + qcalorimeter).

The heat gained by the calorimeter water, qwater, depends on the mass of water, the specific heat of water, Cpand ΔT, while the heat gained by the calorimeter, qcalorimeter, depends on the heat capacity, C and ΔT.

DATA : 1, 2, 3

[tex]T_i(^oC)[/tex] 24.2, 24.0 , 23.2

[tex]T_f(^oC)[/tex] 38.2, 37.8 , 36.6

Mass (g) 70.001 , 70.008 , 70.271

Explanation:

[tex]Q=mc\Delta T[/tex]

Where:

Q = heat absorbed or heat lost

c = specific heat of substance

m = Mass of the substance

ΔT = change in temperature of the substance

Trail 1

Heat absorbed by the water = [tex]Q_1[/tex]

Mass of water ,m,= 70.001 g

Specific heat water = c = 4.184 J/g°C

ΔT = [tex]T_f-T_i=38.2 ^oC - 24.2^oC = 14^oC[/tex]

[tex]Q_2=70.001 g\times 4.184 J/g^oC\times 14^oC=4,100.38 J[/tex]

Heat absorbed by the water is 4,100.38 J.

Trail 2

Heat absorbed by the water = [tex]Q_2[/tex]

Mass of water ,m,= 70.008 g

Specific heat water = c = 4.184 J/g°C

ΔT = [tex]T_f-T_i=37.8 ^oC - 24.0^oC = 14.2^oC[/tex]

[tex]Q_2=70.008 g\times 4.184 J/g^oC\times 14.2^oC=4,042.21 J[/tex]

Heat absorbed by the water is 4,042.21 J.

Trail 3

Heat absorbed by the water = [tex]Q_3[/tex]

Mass of water ,m,= 70.271 g

Specific heat water = c = 4.184 J/g°C

ΔT = [tex]T_f-T_i=36.6^oC - 23.2^oC = 13.4^oC[/tex]

[tex]Q_3=70.271 g\times 4.184 J/g^oC\times 13.18^oC=3939.78 J[/tex]

Heat absorbed by the water is 3,939.78 J.