What is the final “Theoretical Final Temperature? (a) Mass of the Calorimeter (t

ID: 636724 • Letter: W

Question

What is the final “Theoretical Final Temperature?

(a) Mass of the Calorimeter (two empty cups) in g: 3.6 g

(b) Mass of the cold water and the calorimeter in g: 70.30 g

(d) Temp. of the Cold Water in C:9.32 °C

(e) Temp. of the Hot Water before mixing with the cold in C:54.14°C

(f) Mass of the Cold & hot water mixed AND the calorimeter in g: 181.6 g

(g) Calculated Mass of the Hot water in g [(f)-(b)]:181.6 - 66.7= 114.9 g

(h) Temperature of the mixed hot & cold water in C: 38.63 °C

(i) Calculated Heat Lost by the Hot Water

M=114.90 g

C= 4.18J/gC

? T = 38.63 °C – 54.14 °C = -15.51 °C.

Q= (114.90) (4.18J/gC) (-15.51°C.) = -7.5x 10^3J

(j) Calculated Heat Gained by the Cold Water in J

M=66.70 g

C= 4.18J/gC

? T = 38.63 °C – 9.32 °C = 29.31°C.

Q=(66.70) (4.18J/gC) (29.31°C.) = 8.2x 10^3J

What is the final “Theoretical Final Temperature?

(i)Calculated Heat loss by hot water

M1= 114.90 g

C= 4.18 J/gC

?T1= -15.51

Q1= (114.90g) (4.18 J/gC) (-15.51 ) = -7.5 * 103 J (j) Calculated Heat gained by cold water in J

M2=66.70 g

C= 4.18 J/gC

?T2= 29.31

Q2= (66.70g) (4.18 J/gC) (29.31 ) = 8.2 * 103 J

Also,

?T= ?T1?T2

= (-15.51 – 29.31) °C

= -44.82°C

Therefore,

Theoretical Final Temperature= Temperature of mixture of hot and cold water + ?T

= [38.63 + (-44.82)] °C

= -6.19 °C

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