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* c_1* (T_f-T_1)=-m_2* c_2* (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)* (4.184J/g^oC)* (35.0-60.2)^oC=-(108.7g)* c_2* (35.0-19.3)^oC`

`c_2=6.72J/g^oC`

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