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31 March, 14:08

125.0 g of an unknown substance is heated to 97.0°C. It is then placed in a calorimeter than contains 250g of water with an initial temperature of 20.0°C.

The final temperature reached by the calorimeter is 23.5°C.

What is the specific heat of the unknown substance? The specific heat of water is 4.18 J / (°C * g)

A) 0.285 J / (°C * g)

B) 0.398 J / (°C * g)

C) 0.729 J / (°C * g)

D) 1.24 J / (°C * g)

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Answers (1)
  1. 31 March, 17:30
    0
    The heat released by the substance in the calorimeter is equal to the heat absorbed by water which results to the decrease and increase in temperature, respectively.

    We use m Cp ΔT to balance the heat involved

    (m Cp ΔT) subs in calorimeter = (m Cp ΔT) water

    125 g * Cp * (97.0-23.5) C = 250 g * (4.18 J/C g) * (23.5-20)

    Cp = 0.398 J/Cg

    Answer is B
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