Question

A metal sample weighing 129.00 grams and at a temperature of 97.8 degrees Celsius was placed in 45.00 grams of water in a calorimeter at 20.4 degrees Celsius. At equilibrium the temperature of the water and metal was 39.6 degrees Celsius. Calculate the specific heat of the metal. The specific heat of the water is 4.184 J/g/C

Answer 1

Answer : The specific heat of metal is
`0.481J/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 =
`4.184J/g^oC`

`m_1` = mass of metal = 129.00 g

`m_2` = mass of water = 45.00 g

`T_f` = final temperature =
`39.6^oC`

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

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

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

`129.00g* c_1* (39.6-97.8)^oC=-45.00g* 4.184J/g^oC* (39.6-20.4)^oC`

`c_1=0.481J/g^oC`

Therefore, the specific heat of metal is
`0.481J/g^oC`.