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.

$q_1=-q_2$

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

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

$1000g\times c_1\times (26.3-150)^oC=-1000g\times 4.18J/g^oC\times (26.3-25)^oC$

$c_1=0.0439J/g^oC$

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

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

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