Thermochemistry problem

Question

A 75g piece of metal is removed from a pot of boiling water and placed into a calorimeter with 100mL of water that has an initial temperature of 20.8°C. After the metal is put in the water, the new temperature of the water is 24°C. Complete the following table: Specific heat of water = 4.18J/g°C

Answer 1

According to the law of conservation of energy, the heat LOST by the hot metal must equal the heat GAINED by the cooler water. This can be expressed in mathematical form as follows:

-mass metal x sp.heat metal x ∆T metal = mass water x sp.heat water x ∆T water

heat gained by water = q = (100 g)(4.18 J/gº)(24 - 20.8) = (418)(3.2) assuming a density of water = 1g/ml

heat gained by water = 1338 J

heat lost by metal = 1338 J

Initial temperature of metal = 100º (boiling point of water)

We can solve for specific heat of metal as follows:

q = mc∆T

1338 J = (75 g)(c)(100 -24) = (75g)(c)(76)

c = specific heat of metal = 0.235 J/gº