Question

Take a calorimeter that has a specific heat of 40J/K and has a 30 gram piece of ice at (-10 degrees Celsius). What mass of an Iron (Fe) sample that had been heated to 300 degrees Celsius would be required to heat the whole system to 105 degrees Celsius?

Answer 1

Answer

1094g

Explanation

i) Heat required to raise the temperature of caloriemeter (q₁)

q₁ = ∆T × C

= 115K × 40J/K

= 4600J

ii) Heat required to raise the temperature of ice from -10℃ to 0℃ (q₂)

q₂ = m × ∆T × C

= 30g × 10 K × 2.108J /g K

= 632.4J

iii) Heat required to melt ice (q₃)

q₃ = ∆H_fus × n

= 6010J /mol× 1.665 mol

= 10007 J

iv) Heat required to raise the temperature of liquid water from 0℃ to 100℃ (q₄)

q₄ = m × ∆T × C

= 30g × 100K × 4.184J/g K

= 12552J

v) Heat required to vaporize liquid water at 100℃(q₅)

q₅ = ∆H_vap × n

= 40660J/mol × 1.665mol

= 67699J

vi) Heat required to raise the temperature of water vapor from 100℃ to 105℃ (q₇)

q₇ = m × ∆T × C

= 30g × 5K × 1.996J/g K

= 299.4J

vii) Heat released when iron cool from 300℃ to 105℃ (q₇)

q₇ = m × ∆T × C

= m × -195K × 0.449J/g K

= - m 87.56 J/g

viii) Mass of iron required

q₇ = -(q₁ + q₂ + q₃ + q₄ + q₅ + q₆)

- m 87.56J = - (4600J + 632.4J + 10007J + 12552J + 67699J + 299.4J )

- m 87.56J/g = - 95790J

m = 1094g