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SashulF
3 months ago
15

A student wishes to determine the heat capacity of a coffee-cup calorimeter. After she mixes 108.7 g of water at 60.2°C with 108

.7 g of water, already in the calorimeter, at 19.3°C, the final temperature of the water is 35.0°C. Calculate the heat capacity of the calorimeter in J/K. Use 4.184 J/g°C as the specific heat of water. Enter to 1 decimal place.
Physics
1 answer:
Ostrovityanka [3.2K]3 months ago
3 0

Answer: The calorimeter's heat capacity is 6.72J/g^oC

Explanation:

This scenario assumes the amount of heat lost by the hot object equals the amount of heat gained by the cold object.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat capacity of water = 4.184J/g^oC

c_2 = specific heat capacity of calorimeter =?

m_1 = mass of water = 108.7 g

m_2 = mass of calorimeter = 108.7 g

T_f = final temperature of the mixture = 35.0^oC

T_1 = initial temperature of the water = 60.2^oC

T_2 = initial temperature of calorimeter = 19.3^oC

Now substituting all provided values into the formula, we obtain

(108.7g)\times (4.184J/g^oC)\times (35.0-60.2)^oC=-(108.7g)\times c_2\times (35.0-19.3)^oC

c_2=6.72J/g^oC

Hence, the heat capacity of the calorimeter is 6.72J/g^oC

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