Question

A 1000g Ni rod, heated to 150. °C was placed in 1.00 kg of water at 25.0 °C. The final temperature of the water was 26.3 °C. What is the molar specific heat of Ni? QHen Gcald, CAH o) 4184 J/goc

Answer 1

Answer : The molar specific heat of Ni is,
`2.576J/mole^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 Ni = ?

`c_1` = specific heat of water =
`4.18J/g^oC`

`m_1` = mass of Ni = 1000 g

`m_2` = mass of water = 1 kg = 1000 g

`T_f` = final temperature of water =
`26.3^oC`

`T_1` = initial temperature of Ni =
`150^oC`

`T_2` = initial temperature of water =
`25^oC`

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

`1000g* c_1* (26.3-150)^oC=-1000g* 4.18J/g^oC* (26.3-25)^oC`

`c_1=0.0439J/g^oC`

Now we have to calculate the molar specific heat of Ni.

`\text{Molar specific heat of Ni}=\text{Specific heat of Ni}* \text{Molar mass of Ni}=(0.0439J/g^oC)* (58.69g/mole)=2.576J/mole^oC`

Therefore, the molar specific heat of Ni is,
`2.576J/mole^oC`