Question

200 g of an unknown metal is heated to 90.0 degrees C, then dropped into 50.0 g of water at 20.0 degrees C in an insulated container. The water temperature rises within a few seconds to 27.7 degrees C, then changes no further. Identify the metal.

Answer 1

Since heat generate by temperature change is given by,

Q = m*C*dT

Given that final temperature is 27.7 C
Now using energy conservation:
Heat released by metal = Heat gained by liquid
Q1 = Q2
Mm*Cm*dT1 = M*C*dT2
Mm = mass of metal = 200 gm

M = mass of water = 50 gm
Cm = specific heat capacity of metal = ?? J/g-C
C = specific heat capacity of water = 4.2 J/g-C
dT1 = 90 - 27.7 = 62.3
dT2 = 27.7 - 20.0 = 7.7
Using these values:
200*Cm*62.3 = 50*4.2*7.7
Cm = 1617/(200*62.3)
Cm = 0.1298 J/g-C

Cm = 1298 J/kg-C

So,from reference table this specific heat capacity belongs to Iridium metal.

Unknown metal is Iridium

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