An electric heating coil is immersed in a beaker of water and turned on. A current of 3.50 A floured through the coil. After the water starts to boil, it is found that 60.0 g of water vaporizes in 4.00 minutes. What is the resistance of the coil? (Iv = 540 cal/g)

Answer 1

energy in = i^2 R * t
i^2 = 12.25
t = 4*60 = 240 seconds
so energy in = 2940 R Joules

energy to boil = 60 g * 540 cal/g
= 32400 calories

32400 calories(.239J/calorie)
=7744 Joules
so
2940 R = 7744
R = 2.63 Ohms