The specific heat capacity of the metal is determined as 0.76 J/g⁰C.
mass = density x volume
mass = 342 ml x 1 g/ml = 342 g
Apply the law of conservation of energy;
Heat lost by the metal = Heat gained by the calorimeter + heat gained by water
(93.4)(85 - 29.3)c = (8.29)(29.3) + (342)(4.186)(29.3 - 26.7)
5,202.38c = 3,965.088
c = 3,965.088/5,202.38
c = 0.76 J/g⁰C
Thus, the specific heat capacity of the metal is determined as 0.76 J/g⁰C.
The complete question is below;
93.4 g of a metal is heated to 85.0 C in a hot water bath, then dropped into a calorimeter containing 342 ml of water (d 1.00 g/mL) at 26.7 . The equilibrium temperature was found to be 29.3 C. Calculate the specific heat of the metal with proper units. (C.cal 8.29 J/c)
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