Question

If 75 g of metal at 100.0∘C is transferred into 80.0 mL of water at 20.0∘C initially. The water is then raised to 25∘C. What is the specific heat of the metal?

Answer 1

Answer : The specific heat of the metal is,
`0.297J/g^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* c_1* (T_f-T_1)=-m_2* c_2* (T_f-T_2)`

where,

`c_1` = specific heat of metal = ?

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

`m_1` = mass of metal = 75 g

`m_2` = mass of water =
`Density* Volume=1.00g/mL* 80.0mL=80.0g`

`T_f` = final temperature of mixture =
`25^oC`

`T_1` = initial temperature of metal =
`100.0.0^oC`

`T_2` = initial temperature of water =
`20.0^oC`

Now put all the given values in the above formula, we get

`(75g)* c_1* (25-100.0)^oC=-[(80.0g)* 4.18J/g^oC* (25-20.0)^oC]`

`c_1=0.297J/g^oC`

Therefore, the specific heat of the metal is,
`0.297J/g^oC`