Asked by Anonymous
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)
Answers
Answered by
Damon
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
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
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