Question

When 100.0 g block of a metal at 600 oC is plunged into 100.0 g of water (specific heat capacity 4.2 J/g.oC) at 30 oC, the final temperature of both the metal and the water is 80 oC. If no heat is lost to the surroundings, what is the specific heat capacity of the metal in J/g.oC? Give your answer to 2 significant figures.

Answer 1

Answer:
`0.40J/g^0C`

Step-by-step explanation:

`heat_(released)=heat_(absorbed)`

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 = 100.0 g

`m_2` = mass of water = 100.0 g

`T_(final)` = final temperature =
`80^0C`

`T_1` = temperature of metal =
`600^oC`

`T_2` = temperature of water =
`30^oC`

`c_1` = specific heat of metal = ?

`c_2` = specific heat of water=
`4.2J/g^0C`

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

`-[100.0* c_1* (80-600)]=[100.0* 4.2* (80-30)]`

`c_1=0.404J/g^0C`

Therefore, the specific heat capacity of metal is
`c_1=0.40J/g^0C`