Suppose you heat a metal object with a mass of 34.9 g to 96.4 oC and transfer it to a calorimeter containing 100.0 g of water at 17.1 oC. The water and metal reach a final temperature of 22.6 oC. What is the specific heat of the metal in J/g-oC

The specific heat capacity of the metal = 0.894 J/g.°C

Explanation:

With the right assumption that the calorimeter is well insulated and no heat is lost to the calorimeter,

Heat lost by the metal = Heat gained by water

Heat lost by the metal = mCΔT

m = mass of metal = 34.9 g

C = specific heat capacity of the metal = ?

ΔT = change in temperature of the metal = (final temperature) - (Initial temperature)

= 96.4 - 22.6 = 73.8°C

Heat lost by metal = 34.9 × C × 73.8 = (2,575.62C) J

Heat gained by water = mCΔT

m = mass of water = 100 g

C = Specific heat capacity of water = 4.186 J/g.°C

ΔT = change in temperature = 22.6 - 17.1 = 5.5°C

Heat gained by water = 100 × 4.186 × 5.5 = 2,302.3 J

Heat lost by the metal = Heat gained by water

2,575.62C = 2,302.3

C = (2302.3/2575.62) = 0.8938818615 = 0.894 J/g.°C

Hope this Helps!!!

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i'm assuming that it's c.) red bank..