Question

If you drop 50g of metal that is initially at 120ºC into 100g of water in a styrofoam cup that is initially at 20ºC and the temperature of the water rises to 30ºC, what is the specific heat of the metal, if you ignore any heat transferred to the cup?

Answer 1

Answer : The specific heat of metal is,
`0.928J/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 = 50 g

`m_2` = mass of water = 100 g

`T_f` = final temperature =
`30^oC`

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

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

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

`50g* c_1* (30-120)^oC=-100g* 4.18J/g^oC* (30-20)^oC`

`c_1=0.928J/g^oC`

Therefore, the specific heat of metal is,
`0.928J/g^oC`