Question

A 32.4 g iron rod, initially at 21.6 ∘C, is submerged into an unknown mass of water at 63.1 ∘C, in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 59.6 ∘C. You may want tWhat is the mass of the water?

Answer 1

Answer: 37.8 grams

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 rod = 32.4 g

`m_2` = mass of water = ?

`T_(final)` = final temperature =
`59.6^oC=332.6K`

`T_1` = temperature of iron rod =
`21.6^oC=294.6K`

`T_2` = temperature of water =
`63.1^oC=336.1K`

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

`c_2` = specific heat of water=
`4.18J/g^0C`

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

`32.4* 0.450* (332.6-294.6)=-[m_2* 4.18* (332.6-336.1)]`

`m_2=37.8g`

Therefore, the mass of the water is 37.8 grams