Question

The specific heat of mercury is 0.138 J/g Co . If 452g of mercury at 85.0 Co are placed in 145g of water at 23.0 Co , what will be the final temperature for both the mercury and the water?

Answer 1

Answer: The final temperature of the mixture will be
`28.5^0C`

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 mercury = 425 g

`m_2` = mass of water = 145 g

`T_(final)` = final temperature = ?

`T_1` = temperature of mercury =
`85.0^oC`

`T_2` = temperature of water =
`23.0^oC`

`c_1` = specific heat of mercury =
`0.140J/g^0C`

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

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

`-425* 0.140* (T_(final)-85.0)=[145* 4.184* (T_(final)-23.0)]`

`T_(final)=28.5^0C`

Therefore, the final temperature of the mixture will be
`28.5^0C`