Question

When a 40.0 g sample of a metal at 25.00 °C is added to 65.0 g of water at 100.00 °C, the final temperature of both the water and metal is 93.27 °C. The specific heat of water is 4.184 J/(g×C). What is the specific heat of the metal?

Answer 1

Answer: Specific heat of the metal is
`0.670J/g^0C`

Step-by-step explanation:

`heat_(absorbed)=heat_(released)`

As we know that,

`Q=m* c* \Delta T=m* c* (T_(final)-T_(initial))`

`m_1* c_1* (T_(final)-T_1)=-[m_2* c_2* (T_(final)-T_2)]` .................(1)

where,

q = heat absorbed or released

`m_1` = mass of metal = 40.0 g

`m_2` = mass of water = 65.0 g

`T_(final)` = final temperature =
`93.27^oC=(93.27+273)K=366.27K`

`T_1` = temperature of metal =
`25^oC=(25+273)K=298K`

`T_2` = temperature of water =
`100^oC=(100+273)K=373K`

`c_1` = specific heat of metal = ?

`c_2` = specific heat of water=
`4.184J/g^0C`

Now put all the given values in equation (1), we get

`40.0* c_1* (366.27-298)=-[65.0* 4.184* (366.27-373)]`

`c_1=0.670`

Therefore, the specific heat of the metal is
`0.670J/g^0C`