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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?

0.171
0.343
1.717
3.433

Respuesta :

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The piece of unknown metal is in thermal equilibrium with water such that Q of metal is equal to Q of the water. We write this equality as follows:

-Qm = Qw
Mass of metal (Cm)(ΔT) = Mass of water (Cw) (ΔT)

where C is the specific heat capacities of the materials.

We calculate as follows:

-(Mass of metal (Cm)(ΔT)) = Mass of water (Cw) (ΔT)
-68.6 (Cm)(52.1 - 100) = 42 (4.184) (52.1 - 20)
Cm = 1.717 -----> OPTION C

Answer:

Option (d)

Explanation:

Let c be the specific heat of the metal.

By use of principle of caloriemetry,

Heat lost by the hot body = Heat gained by the cold body

Heat lost by metal = Heat gained by water

mass of metal x specific heat of metal x fall in temperature of metal = mass of water x specific heat of water x rise in temperature of water

68.6 x c x (100 - 52.1) = 84 x 4.184 x (52.1 - 20)

3285.94 x c = 11281.7376

c = 3.433 J/g C

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