Answer : The specific heat of unknown sample is, 
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]](https://tex.z-dn.net/?f=q_1%3D-%5Bq_2%2Bq_3%5D)
![m_1\times c_1\times (T_f-T_1)=-[m_2\times c_2\times (T_f-T_2)+m_3\times c_3\times (T_f-T_2)]](https://tex.z-dn.net/?f=m_1%5Ctimes%20c_1%5Ctimes%20%28T_f-T_1%29%3D-%5Bm_2%5Ctimes%20c_2%5Ctimes%20%28T_f-T_2%29%2Bm_3%5Ctimes%20c_3%5Ctimes%20%28T_f-T_2%29%5D)
where,
= specific heat of unknown sample = ?
= specific heat of water = 
= specific heat of copper = 
= mass of unknown sample = 72.0 g = 0.072 kg
= mass of water = 203 g = 0.203 kg
= mass of copper = 187 g = 0.187 kg
= final temperature of calorimeter = 
= initial temperature of unknown sample = 
= initial temperature of water and copper = 
Now put all the given values in the above formula, we get
![0.072kg\times c_1\times (39.4-80.0)^oC=-[(0.203kg\times 4186J/kg^oC\times (39.4-11.0)^oC)+(0.187kg\times 390J/kg^oC\times (39.4-11.0)^oC)]](https://tex.z-dn.net/?f=0.072kg%5Ctimes%20c_1%5Ctimes%20%2839.4-80.0%29%5EoC%3D-%5B%280.203kg%5Ctimes%204186J%2Fkg%5EoC%5Ctimes%20%2839.4-11.0%29%5EoC%29%2B%280.187kg%5Ctimes%20390J%2Fkg%5EoC%5Ctimes%20%2839.4-11.0%29%5EoC%29%5D)
Therefore, the specific heat of unknown sample is,
