Question

Imagine that you have a cup of water containing 88.4 g. The specific heat of water is 4.19 J/g

Answer 1

Heat gained by water = m.c.del(T) [m=mass of water, c= specific heat capacity of water, del(T)= change in temperature]

Here, m =88.4 gm , c= 4.19 J/g C ; T= 58.2-20 = 38.2

Now, Heat supplied by immersion rod = P*t [P=Power of the rod; t= time for which it works]

Heat supplied by rod = heat taken by rod => 100 *t = 88.4*4.19*38.2 = 141.49 sec

For half of the water to vaporize, heat supplied = m/2 * L [L=latent heat of vaporization of water]

= 88.4/2 g* 2.26*10^3 J/g

Heat supplied by rod = 100*t

100*t = 88.4/2 * 2.26*10^3

t=998.92 sec