Question

A 360. g iron rod is placed into 750.0 g of water at 22.5°C. The water temperature rises to 46.7°C. What was the initial temperature of the iron rod?

Answer 1

Answer: The initial temperature of the iron was $515^0C$

Explanation:

$heat_{absorbed}=heat_{released}$

As we know that,  

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$         .................(1)

where,

q = heat absorbed or released

$m_1$ = mass of iron = 360 g

$m_2$ = mass of water = 750 g

$T_{final}$ = final temperature = $46.7^0C$

$T_1$ = temperature of iron = ?

$T_2$ = temperature of water = $22.5^oC$

$c_1$ = specific heat of iron = $0.450J/g^0C$

$c_2$ = specific heat of water= $4.184J/g^0C$

Now put all the given values in equation (1), we get

$-360\times 0.450\times (46.7-x)=[750\times 4.184\times (46.7-22.5)]$

$T_i=515^0C$

Therefore, the initial temperature of the iron was $515^0C$