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#67090 to #66874

GeorgeBush (02/01/2013) []
There is a house hold heater that operates at 4 V and at 35 Ω and is used to heat up 15g of copper wire. The specific heat capacity of copper is 24.440 J/mol/K. How much time is required to increase the temperature from 25˚C to 69˚C?
#67132 to #67117

GeorgeBush (02/01/2013) []
I'll try
Most of what that equation is lies in pluggin in information after you have the prompt. To find the power (P), you have to divide your volts (squared) by the resistance, P =V^2/R. So plug the numbers we know in:
P = (4)^2/(35)
P = (16)/(35), coming finally to P = .457 J/s
Now we need to find the equation by circuitry that will calculate power, which is q=mCsΔT*, with a note that "s" in that equation is meant to be subscript.
Now we plug in the numbers we know into the equation.
q=(15)(24.440)(6925) where 15(g) is the total weight of the copper wire, 24.440(J/mol/K) is the cumulative heat capacity for said wire, and 6925 is the difference in temperature. So,
q= (15)(24.440)(6925) = 16130.4 J
So now we're left with 2 numbers, one tells us how many joules we must add to heat the metal, and the other tells us per second how many joules of energy the heater gives out. Now we just divide by the number of seconds. So,
(16130.4 J)/(.457 J/s) = 35296.3 seconds. 35296.3 seconds/60 gives us 588.27120 minutes. That, in turn, divided by 60 figures out to be 9.804530 hours.
Most of what that equation is lies in pluggin in information after you have the prompt. To find the power (P), you have to divide your volts (squared) by the resistance, P =V^2/R. So plug the numbers we know in:
P = (4)^2/(35)
P = (16)/(35), coming finally to P = .457 J/s
Now we need to find the equation by circuitry that will calculate power, which is q=mCsΔT*, with a note that "s" in that equation is meant to be subscript.
Now we plug in the numbers we know into the equation.
q=(15)(24.440)(6925) where 15(g) is the total weight of the copper wire, 24.440(J/mol/K) is the cumulative heat capacity for said wire, and 6925 is the difference in temperature. So,
q= (15)(24.440)(6925) = 16130.4 J
So now we're left with 2 numbers, one tells us how many joules we must add to heat the metal, and the other tells us per second how many joules of energy the heater gives out. Now we just divide by the number of seconds. So,
(16130.4 J)/(.457 J/s) = 35296.3 seconds. 35296.3 seconds/60 gives us 588.27120 minutes. That, in turn, divided by 60 figures out to be 9.804530 hours.