Question

An electrical heating coil is immersed in 4.1kg of water at 20 degrees C. The coil, which has a resistance of 250 ?, warms the water to 34 degrees C in 14min .

Part A

What is the potential difference at which the coil operates?

Express your answer using two significant figures.

V =

kV

Answer 1

A) P resistor = V^2/R

P needed = c m dT/t

V^2/250 = 4186*4.1*(34-20)/(14*60)

V=267 V = 0.27 kV

Answer 2

heat energy =mc(T2-T1)=4.1*4190*(34-20) =240506

time t = 14 min =14*60 = 840 sec

power =V^2/R

240506/840 = V^2/250

V =267.54 volts = 0.267 kV

Answer 3

Heat required to heat water to 34? from 20? =mC?T=4.1*4.18*14

=239.9 kJ

Power= Energy/time=239.9/14*60=0.2856 kW=285.6 W

Power=V²/R

285.6=V²/250

V=267.2 V=0.267 kV

Answer 4

We use the formula

me(n-n)=를,

m = 4.1kg
T1 = 20 + 273 = 293K
T2 = 34 + 273 = 307K
t = 14*60 = 840s
R = 250 ?,
c = 4190 J / kg.K
Substitute all the given values in the above mention formulawe get the value of ?.

? = 267.54 V = 0.27 kV