Question

A laboratory technician drops a 72.0 g sample of unknown solid material, at a temperature of 80.0°C, into a calorimeter. The calorimeter can, initially at 11.0°C, is made of 187 g of copper and contains 203 g of water. The final temperature of the calorimeter can and contents is 39.4°C. What is the specific heat of the unknown sample? Please give your answer in units of J/kg.°C.

Answer 1

Answer : The specific heat of unknown sample is,
`8748.78J/kg^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+q_3]`

`m_1* c_1* (T_f-T_1)=-[m_2* c_2* (T_f-T_2)+m_3* c_3* (T_f-T_2)]`

where,

`c_1` = specific heat of unknown sample = ?

`c_2` = specific heat of water =
`4186J/kg^oC`

`c_3` = specific heat of copper =
`390J/kg^oC`

`m_1` = mass of unknown sample = 72.0 g = 0.072 kg

`m_2` = mass of water = 203 g = 0.203 kg

`m_2` = mass of copper = 187 g = 0.187 kg

`T_f` = final temperature of calorimeter =
`39.4^oC`

`T_1` = initial temperature of unknown sample =
`80.0^oC`

`T_2` = initial temperature of water and copper =
`11.0^oC`

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

`0.072kg* c_1* (39.4-80.0)^oC=-[(0.203kg* 4186J/kg^oC* (39.4-11.0)^oC)+(0.187kg* 390J/kg^oC* (39.4-11.0)^oC)]`

`c_1=8748.78J/kg^oC`

Therefore, the specific heat of unknown sample is,
`8748.78J/kg^oC`