Jay0801
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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?

Respuesta :

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: The specific heat of the metal comes out to be [tex]3.423J/g^oC[/tex]

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

[tex]q=mc\Delta T[/tex]

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,

[tex](mc\Delta T)_m=-(mc\Delta T)_w[/tex]     ....(1)

  • For metal:

[tex]T_2=52.1^oC[/tex]

[tex]T_1=100^oC[/tex]

[tex]\Delta T=T_2-T-1=(52.1-100)^oC=-47.9^oC[/tex]

m = 68.6 g

c = ?

  • For Water:

[tex]T_2=52.1^oC[/tex]

[tex]T_1=20^oC[/tex]

[tex]\Delta T=T_2-T-1=(52.1-20)^oC=32.1^oC[/tex]

m = 84 g

[tex]c=4.184J/g^oC[/tex]

Putting this in equation 1, we get

[tex]68.6g\times c_m\times (-47.9^oC)=84g\times 4.184J/g^oC\times 32.1^oC[/tex]

[tex]c_m=3.423j/g^oC[/tex]

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