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`

Step-by-step explanation:

`heat_(absorbed)=heat_(released)`

As we know that,

`Q=m* c* \Delta T=m* c* (T_(final)-T_(initial))`

`m_1* c_1* (T_(final)-T_1)=-[m_2* c_2* (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* 0.450* (46.7-x)=[750* 4.184* (46.7-22.5)]`

`T_i=515^0C`

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