Question

A piece of unknown metal with mass 68.6 g is heated to an initial temperature of 100 °C and dropped into 84 g of water (with an initial temperature of 20 °C) in a calorimeter. The final temperature of the system is 52.1°C. The specific heat of water is 4.184 J/g*⁰C. What is the specific heat of the metal?

Answer 1

52.1-4.184=47.916 You then either add or subtract this from 68.6 the first temperature of the metal. This will give you your total.

Answer 2

Answer: The specific heat of the metal comes out to be
`3.423J/g^oC`

Explanation: Heat absorbed or heat lost, q is written as:

`q=mc\Delta T`

We are given that an unknown metal is dropped in water. So, in this process:

Heat lost by the metal = Heat gained by the water

Mathematically,

`(mc\Delta T)_m=-(mc\Delta T)_w` ....(1)

• For metal:

`T_2=52.1^oC`

`T_1=100^oC`

`\Delta T=T_2-T-1=(52.1-100)^oC=-47.9^oC`

m = 68.6 g

c = ?

• For Water:

`T_2=52.1^oC`

`T_1=20^oC`

`\Delta T=T_2-T-1=(52.1-20)^oC=32.1^oC`

m = 84 g

`c=4.184J/g^oC`

Putting this in equation 1, we get

`68.6g* c_m* (-47.9^oC)=84g* 4.184J/g^oC* 32.1^oC`

`c_m=3.423j/g^oC`