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20.45g of a substance, initially at 97.20C was poured into 10.08 g of water in a...

20.45g of a substance, initially at 97.20C was poured into 10.08 g of water in a calorimeter. The initial temperature of the water and the calorimeter was 21.37C. The calorimeter has a heat capacity of 9.76 J/K. The temperature after mixing, Tf, was determined to be 32.66C. Calculate the specific heat capacity of the substance.
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Answer #1

Answer

0.444 J/g ℃

Explanation

Heat change of the substance ( qs) = -( heat change of water(qw) + heat change of calorimeter(qc))

qs = m × ∆T × Cs

= 20.45g × ( 32.66℃ - 97.20℃) × Cs

= -1319.84g℃ × Cs

qw =m × ∆T × Cs

=10.08g × ( 32.66℃ - 21.37℃) × 4.184J/g℃

= 476.15J

qc = ∆T × Cs

= ( 32.66℃ - 21.37℃ ) × 9.76J/℃

= 110.19J

qs = - ( qw + qc)

-1319.84 g℃ × Cs = - ( 476.15J + 110.19J)

-1319.84 g℃ × Cs = -586.34J

Cs = 0.444J/g℃

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