Question
Question: One end of the uniform rod of length 1 m is placed in boiling water while its other end is placed in...
One end of the uniform rod of length 1 m is placed in boiling water while its other end is placed in melting ice. A point P on the rod is maintained at a constant temperature of 800∘C. The mass of steam produced per second is equal to the mass of ice melted per second. If the specific latent heat of steam is 7 times the specific heat of ice, the distance of P from the steam chamber must be
a) 91m
b) 81m
c) 71m
d) 101m
Solution
Let us assume that the point P will lie at distance x m. Now, by using the formula Q=lKAΔT, find the rate of heat for both ice and water. Also, we know that: Q=m×s, where m is mass and s is specific heat of fusion for ice and specific heat of evaporation for water. So, get the value of specific heats in terms of x. As it is mentioned, the specific latent heat of steam is 7 times the specific heat of ice. So, form a single equation in one variable, and solve for x.
Complete step by step answer:
The heat rate by using the formula Q=lKAΔT=m×s
For ice:
Q1=xKA(800−0)=m×sice......(1)
For water:
Q2=(l−x)KA(800−100)=m×swater......(2)
Divide equation (1) by (2), we get: