Asample of iron metal absorbs 4.56 j of heat, upon which the temperature of the sample increased from 29 °c to 36 °c. which of the following is the mass of the sample? cs of iron is 0.46 j/g °c.
a) 1.22 g
b) 1.42 g
c) 4.12 g
d) 3.22 g

b) 1.42 g

Explanation:

Data Given:

Amount of Heat = 4.56 J

initial temperature = 29 °C

Final temperature = 36 °C

Cs of iron = 0.46 J/g °C

mass of sample = ?

Solution:

Formula used

              Q = Cs.m.ΔT

rearrange the above equation to calculate the mass of iron sample

             m =  Q / Cs.ΔT   . . . . . (1)

Where:

Q = amount of heat

Cs = specific heat of iron = 0.46 J/g °C

m = mass

ΔT = (t2 - t1) = Change in temperature

First we have to find ΔT

                   ΔT = (t2 - t1)

                   ΔT = (36 °C - 29°C )

                   ΔT = 7 °C

Put values in above equation 1

                 m = 4.56 J / 0.46 (J/g °C) x 7 °C

                 m =4.56 J / 3.22 (J/g)

                 m =  1.42 g                  

So option b is correct = 1.42 g

the volume increases