Question

What is the specific heat capacity of an unknown metal if 55.00 g of the metal initially at 90°C is placed in 75mL of water with an initial temperature of 25°C and the final temperature of the system is 35°C?

Answer 1

The specific heat capacity of the unknown metal is 1.04 J/g°C.

Step-by-step explanation:

Let the specific heat capacity of the unknown metal be
`x`.

Given:

Mass of the metal is,
`m=55\ g`

Initial temperature of the metal is,
`T_i=90\ \°C`

Volume of water is,
`V=75\ ml`

Specific heat capacity of water is,
`c_w=4.186\ J/g\°C`

Initial temperature of water is,
`T_(wi)=25\ \°C`

Final temperature of the system is,
`T=35\ \°C`

We know that density of water is equal to 1 g/ml.

Mass is given as the product of density and volume.

Therefore, mass of water is given as:

`m_w=1* 75=75\ g`

Now, fall of temperature of the unknown metal is given as:

`\Delta T_m=T_i-T=90-35=55\ \°C`

Rise of temperature of water is given as:

`\Delta T_w=T-T_(wi)=35-25=10\ \°C`

Now, as per conservation of energy,

Heat lost by metal = Heat gained by water

⇒
`mx\Delta T_m=m_wc_w\Delta T_w`

Plug in all the given values and solve for
`x`. This gives,

`55* x* 55=75* 4.186* 10\\3025x=3139.5\\x=(3139.5)/(3025)=1.04\ J/g\°C`

Therefore, the specific heat capacity of the unknown metal is 1.04 J/g°C.