metal sample weighing 45.2 g and at a temperature of 100.0°C was placed in 38.6

Question

metal sample weighing 45.2 g and at a temperature of 100.0°C was placed in 38.6 g of water in a calorimeter 33.0°C. r at 25.2°C. At equilibrium, the temperature of the water and metal was a. What was AT for the water? ATo b. What was AT for the metal? oC c. Taking the specific heat capacity of water to be 4.184 J/g°C, how much heat was transferred to the water? d. Calculate the specific heat capaciry of the metal, using Equation 4. e. Calculate the approximate molar mass of the metal, using Equation 5 g/mol

Explanation / Answer

Ans 1 :

a) The change in temperature of water = 33.0 - 25.2

= 7.8oC

b) The change in temperature for metal = 100 - 25.2

= 74.8oC

c) heat = m. c . delta T

putting all the values :

Q = 38.6 x 4.184 x 7.8

= 1259.7 J

d) For metal :

c = 1259.7 / (45.2 x 74.8)

= 0.3726 J /goC

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