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.

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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

queen_honey_blossom answered 2 years ago

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