A 110.0-g sample of a grey-colored, unknown, pure metal was heated to 92.0 degre

Question

A 110.0-g sample of a grey-colored, unknown, pure metal was heated to 92.0 degree C and put into a coffee-cup calorimeter containing 75.0 g of water at 21.0 degree. When the heated metal was put into the water, the temperature of the water rose to a final temperature of 24.2 degree C. The specific heat of water is 4.184 j/g degree C. (a) What is the specific heat of the metal? (b) Is it possible that the metal is either iron or lead? Explain. has a specific heat of 0.47 J/g degree and has a specific heat of 0.13 J/g degree C. The metal could possibly be but further tests would be needed to determine this. The metal cannot be .

Explanation / Answer

a) We know, Heat gained by the water = heat lost by the metal

Q = mcT

where, Q= quantity of heat

m = mass in g

c = specific heat in J/g0C (kJ/kg K), and

T = temp change.

Heat gained by water =75.0*4.184*(24.2 - 21) = 1004 J = Heat lost by metal

So,

1004 = 110.0*c*(92.0-24.2) [for metal]

So, c= 1004/(110.0×67.8) = 0.135 J/g0C

c for lead id given as 0.13 J/g0C. So, the metal is lead.

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