Question

When 5.000 g of NaHCO3 is added to 600.0 g of water in a ‘coffee cup calorimeter’, the temperature decreases from 22.35°C to 18.94°C.

a) Calculate qcalorimeter, in joules, assuming the specific heat capacity of the solution is the same as pure water: 4.186 Jg-1K-1.

Report your value to 4 significant figures, and do not include units in your answer.

Answer 1

Answer is: qcalorimeter is 8635.9273 J.

ΔT= 22.35°C - 18.94°C.

ΔT = 3.41°C = 3.41 K; temperature change.  

m(H₂O) = 600.0 g; mass of water.  

mr = 605.0; mass of solution.

cp= 4.186 J/g·K, specific heat capacity of the solution.  

qcal = mr · ΔT· cp.  

qcal = 605 g · 3.41 K · 4.186 J/g·K.  

qcal = 8635.9273 J; amount of heat absorbed.

Answer 2

Amount of heat absorbed of calorimeter when 5g of NaHCO₃ is added to 600g of water in a ‘coffee cup calorimeter' is 700,396.4968 J.

How do we calculate required heat?

Amount of absorbed heat will be calculated as:
Q = mcΔT, where

ΔT = change in temperature = 22.35°C - 18.94°C = 3.41°C = 276.56 K

m = mass of solution = (5g + 600g) = 605g

c = specific heat of calorimeter = 4.186 J/g.K

On putting these values on the above equation, we get

Q = (605)(4.186)(276.56)

Q = 700,396.4968 J

Hence heat of calorimeter is 700,396.4968 J.

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