In the above problem if heat is supplied at a constant rate of `q= 10 cal//min`, then plot temperature versus time graph.

Answer

A

Explanation

From A to B
(i) Temperature of ice will increase from `-15^@C to 0^@C`.
(ii) `t_(AB) = ("Total heat required")/("Heat supplied per minute") = (Q_1)/q`
`= (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) `t_(BC) = (Q_2)/q = (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^@ to 100^@C`.
(ii) `t_(CD) = Q_3/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) = Q_4/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.
The corresponding graph is as shown in fig.