In the above problem if heat is supplied at a constant rate of q= 10 cal//min, then plot temperature versus time graph.
Answers
Total time taken is t(total) = 581.56 minutes.
Explanation:
From A to B
(i) Temperature of ice will increase from −15∘C → 0∘C.
(ii) t AB=Total heat required / Heat supplied per minute
t AB = Q1 / q
t AB = 63.6 / 10 = 6.36 min
(iii) Between A and B we will get only ice.
From B to C
(i) Temperature of (ice + water) mixture will remain constant at 0∘C.
(ii) tBC=Q2q = 640 / 10 = 64 min
t(Total) = t(AB) + t(BC) = 70.36 min
(iii) Between B and C we will get both ice and water.
From C to D
(i) Temperature of water increases from 0∘ → 100∘C.
(ii) t(CD) = Q3 / q = 800 / 10 = 80 min
t(Total) = t(AB) + t(BC) + t(CD) = 150.36 min
(iii) Between C and D we will get only water.
From D to E
(i) Temperature of (water+steam) mixture will remain constant at 100∘C.
(ii) t(DE) = Q4 / q =4312 / 10=431.2 min
t(Total) =t(AB) + t(BC) + t(CD) + t(DE) = 581.56 min.
(iii) Between D and A we will get both water and steam.