Question

. 125g of water has an initial temperature of 25.6°C, and is heated by 50.0g of a metal

which has been heated to 115.0°C. The metal heats the water so that both the metal
and the water reach a final temperature of 29.3°C. Calculate the specific heat of the
metal.​

Answer 1

Answer : The specific heat of the metal is,
`0.451J/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 = 50.0 g

`m_2` = mass of water = 125 g

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

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

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

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

`(50.0g)* c_1* (29.3-115.0)^oC=-[(125g)* 4.18J/g^oC* (29.3-25.6)^oC]`

`c_1=0.451J/g^oC`

Therefore, the specific heat of the metal is,
`0.451J/g^oC`