Skip to content
Example · Example 4

Q.Calculate the total quantity of heat required to convert 500 g500\ \text{g} of ice at −10 ∘C-10\,^\circ\text{C} completely into steam at 100 ∘C100\,^\circ\text{C} at normal atmospheric pressure. Use cice=2100 J kg−1K−1c_{\text{ice}} = 2100\ \text{J kg}^{-1}\text{K}^{-1}, cwater=4200 J kg−1K−1c_{\text{water}} = 4200\ \text{J kg}^{-1}\text{K}^{-1}, Lf=3.34×105 J kg−1L_f = 3.34\times10^{5}\ \text{J kg}^{-1}, and Lv=2.26×106 J kg−1L_v = 2.26\times10^{6}\ \text{J kg}^{-1}.

West Bengal WbchseTextbookSubjectiveImportance★★★★★est
17% · 4/23 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

The process has four distinct stages, and the heat needed for each must be added, since each stage happens one after another. With m=0.5 kgm = 0.5\ \text{kg}:

Stage 1 -- heating ice from −10 ∘C-10\,^\circ\text{C} to 0 ∘C0\,^\circ\text{C}:

Q1=mcice ΔT=0.5×2100×10=10,500 JQ_1 = mc_{\text{ice}}\,\Delta T = 0.5\times2100\times10 = 10{,}500\ \text{J}

Stage 2 -- melting the ice at 0 ∘C0\,^\circ\text{C} (constant temperature):

Q2=mLf=0.5×3.34×105=167,000 JQ_2 = mL_f = 0.5\times3.34\times10^{5} = 167{,}000\ \text{J}

Stage 3 -- heating the resulting water from 0 ∘C0\,^\circ\text{C} to 100 ∘C100\,^\circ\text{C}:

Q3=mcwater ΔT=0.5×4200×100=210,000 JQ_3 = mc_{\text{water}}\,\Delta T = 0.5\times4200\times100 = 210{,}000\ \text{J} …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.