Question

If a 95.0 gram sample of metal at 100.0 oC is placed in 50.0 g of water with an initial temperature of 22.5oC and the final temperature of the system is 48.5oC, what is the specific heat of the metal? Group of answer choices

Answer 1

Answer : The specific heat of the metal is,
`1.11J/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 = 95.0g

`m_2` = mass of water = 50.0 g

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

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

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

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

`(95.0g)* c_1* (48.5-100.0)^oC=-[(50.0g)* 4.18J/g^oC* (48.5-22.5)^oC]`

`c_1=1.11J/g^oC`

Therefore, the specific heat of the metal is,
`1.11J/g^oC`