Calculate the approx, mate molar mass. of the metal. using Ration 5. Calculate t

Question

Calculate the approx, mate molar mass. of the metal. using Ration 5. Calculate the specific heat capacity of the metal, using liquation 4. Taking the specific heat capacity of water to be 4.184 J/g degree C. how much heat was transferred to the water?) What was Delta T of the metal? A metal weighing 45.2 g and at a temperature of 100.0 degree C was placed in 38.6 g of water in a calorimeter at 25.2 degree C. at equilibrium the temperature of the water and metal was 33.0 degree C. What was Delta T for the water? (Delta T = T_ -T_

Explanation / Answer

1. Mass of metal sample = 45.2 g

Temperature of metal sample = 100C

Mass of water = 38.6 g

Temperature of water (the water assumes the temperature of the calorimeter) = 25.2C

Final temperature attained by the metal and water = 33.0C

                         

a) T (water) = Tfinal – Tinitial = (33.0C) – (25.2C) = 7.8C (ans).

b) T (metal) = Tinitial – Tfinal = (100.0C) – (33.0C) = 67C (ans).

c) Specific heat capacity of water = 4.184 J/g.C

Heat transferred to water = (mass of water)*(specific heat capacity of water)*T (water) = (38.6 g)*(4.184 J/gC)*(7.8C) = 1259.7187 J 1259.72 J (ans).

d) As per the principle of thermochemistry, the heat lost by the metal = the heat gained by water.

Therefore, heat lost by metal = 1259.72 J.

Again, heat lost by metal = (mass of metal)*(specific heat capacity of metal)*T (metal)

Therefore,

(45.2 g)*(specific heat capacity of metal)*(67C) = 1259.72 J

===> specific heat capacity of metal = (1259.72 J)/(45.2 g).(67C) = 0.4159 J/g.C 0.416 J/g.C (ans)

e) I need equation 5 to answer this question.

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