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A 0.016-kg piece of iron absorbs 1086.75 joules of heat energy, and its temperature changes from 25°C to 175°C. Calculate the specific heat capacity of iron. Equation : c = Q/mt * 2608.2 J/(Kg*°C) 3477.6 J/(Kg*°C) 452.8 J/(Kg*°C) 8.7 J/(Kg*°C)

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abidemiokin

Answer:

452.8125J/Kg°C

Explanation:

The quantity of eat absorbed by the iron is expressed as;

Q = mc∆t

Q = 1086.75 joules

m is the mass= 0.016kg

c is the specific heat capacity

∆t is the change in temperature = 175-25 = 150°C

Get c;

From the formula;

c = Q/m∆t

c = 1086.75/0.016(150)

c =1086.75/2.4

c = 452.8125J/Kg°C

Hence the specific heat capacity of iron is 452.8125J/Kg°C

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