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50.3g H2O at 60.5C is added to 48.7g H2O 19.7 C to give a mixture at 39.1 C. What is the Heat compactly of the Calorimeter in this experiment.?

Answer :

Final answer:

The heat capacity of the calorimeter in this experiment is 100.6 J/°C.

Explanation:

To find the heat capacity of the calorimeter, we can use the heat exchange formula:

q = m * c * ΔT

where:

q = heat exchanged

m = mass of the substance

c = specific heat capacity

ΔT = change in temperature

Given that the heat exchange is zero (since the calorimeter is isolated), we can set up the following equation:

m1 * c1 * ΔT1 + m2 * c2 * ΔT2 = 0

Solving for the heat capacity of the calorimeter (c2):

c2 = - (m1 * c1 * ΔT1) / (m2 * ΔT2)

Plugging in the given values:

c2 = - [(50.3g * 4.18 J/g°C * (39.1°C - 60.5°C)] / (48.7g * (39.1°C - 19.7°C))

c2 ≈ 100.6 J/°C

The negative sign arises because the heat lost by one substance is equal to the heat gained by the other. This indicates the direction of heat flow.

Learn more about Heat capacity:

brainly.com/question/34660878

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Rewritten by : Jeany