Question

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.

Answer 1

Answer : The heat capacity of the calorimeter is, $6.72J/g^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

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

where,

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

$c_2$ = specific heat of calorimeter = ?

$m_1$ = mass of water = 108.7 g

$m_2$ = mass of calorimeter = 108.7 g

$T_f$ = final temperature of mixture = $35.0^oC$

$T_1$ = initial temperature of water = $60.2^oC$

$T_2$ = initial temperature of calorimeter = $19.3^oC$

Now put all the given values in the above formula, we get

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

$c_2=6.72J/g^oC$

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