Question
In: Chemistry
131 g of metal at 100.0 oC is added to 60.0 mL of water at 26.0...
131 g of metal at 100.0 oC is added to 60.0 mL of water at 26.0 oC. The equilibrium temperature reaches 39.9 oC. The specific heat of water is 4.184 J/g.K. Assuming the heat capacity of the calorimeter is negligible, calculate the specific heat of the metal.
Solutions
Expert Solution
Sol :-
Given volume of water = 60 mL
Density of water = 1.0 g/mL
So,
Mass of water =Density of water x volume of water
= 60 mL x 1.00 g/mL
= 60 g
As,
Heat released by water (q) = Mass of water x specific heat capacity of water x change in temperature
= 60 g x 4.184 J/g.K x (39.9 - 26) K
= 3489.456 J
Because,
Amount of heat gain by metal = Amount of heat released by water
So,
Mass of metal x specific heat capacity of metal x change in temperature = 3489.456 J
Specific heat capacity of metal = 7618.6 J / (Mass of metal x change in temperature)
= 3489.456 J / (131 g x (100-39.9)K)
= 3489.456 J / (131 g x 60.1 K)
= 0.443 J/g.K
| Hence, Specific heat capacity of metal = 0.443 J/g.K |
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