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9 days ago

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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.

1 answer:

Ostrovityanka [942]9 days 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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