Question

What is the specific heat of a substance if a mass of 10.0 kg increases in temperature from 10.0°C to 70.0°C when 2,520 J of heat is applied?
0.00420 J/(gi°C)
0.00661 J/(gi°C)
238 J/(gi°C)
252 J/(gi°C)

Answer 1

We use the expression Heat = mC(Tf-Ti) where m is the mass of the sample, C is the specific heat and Tf, Ti are the final and initial temperature to calculate the problem above. We do as follows:

2520 = (10000)C ( 70-10)
C = 0.00420 J / g °C <------FIRST OPTION 

Answer 2

The Specific heat of the substance when the temperature is increased from 10 to 70.0°C is 0.00420 J/(gi°C)

Specific Heat Capacity

Given Data

• Mass m = 10 kg to gram = 10*1000 = 10000

• T1 = 10.0°C

• T2 = 70.0°C

• Heat Applied Q = 2,520 J

Applying the relation

Q = mcΔT

Substituting our given data into te formula we have

2,520 = 10*c(70-10)

2,520 = 10000*c*60

2,520 = 600000c

Divide both sides by 600000

c = 2,520/600000

c = 0.00420 J/(gi°C)

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