Pleae answer #1 a through e. Thank you ! A metal sample weighing 124.10 g and at

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Pleae answer #1 a through e. Thank you !

A metal sample weighing 124.10 g and at a temperature of 99.3 degree C was placed in 42.87 g of water in a calorimeter at 23.9degreeC. At equilibrium the temperature of the water and metal was 41.6degreeC. What was Deltat for the water? (Deltat =tfinal-tinitial) degreeC What was At for the metal? degreeC How much heat flowed into the water? joules Taking the specific heat of water to be 4.18 J/gdegreeC, calculate the specific heat of the metal, using Equation 3. joules/gdegreeC What is the approximate molar mass of the metal? (Use Eq. 4.) g

Explanation / Answer

a) delta T for water = 41.6 - 23.9 = 17.7 0C i.e the temperature increases

b) delta T for metal = 41.6 - 99.3 = -57.7 0C i.e. the temperature decreases

c) Heat flowed into water = mass of water*specific heat of water*delta T = 42.87*4.18*17.7 = 3171.78 J

d) heat gained by water = heat lost by metal

or, mass of water*specific heat of water*delta T of water = mass of metal*specific heat of metal*delta T of metal

or, 42.87*4.18*17.7 = 124.1*specific heat of metal*57.7

or, specific heat of metal = 0.443 J/g/0C

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