A 21 Ohm resistor is connected across a 120 V source. The resistor is then lowered into an insulated beaker, containing 1.0 L of water at 23 degrees Celsius for 30 seconds. What is the final temperature of the water? Thank you for any help!!!
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The resistor dissipates P = U^2/R = 120^2/21 = 685.7 W = 685.7 J/s
The specific heat of water is c = 4.186 J / g C
The density is 1000 g / L
Q = c m dT
Q= energy added
c = specific heat
m = mass
dT = temperature change
685.7 J/s 30 s = 4.186 J / g C 1000 g dT
dT = 4.91 C
The final T will be 23 + 4.91 = 27.91 C
Note the use of units, you are less likely to make mistakes if you also pay attention to what units are
used.
I hoep the leads to the resistor are insulated or you will short out your 120V source ;)
The specific heat of water is c = 4.186 J / g C
The density is 1000 g / L
Q = c m dT
Q= energy added
c = specific heat
m = mass
dT = temperature change
685.7 J/s 30 s = 4.186 J / g C 1000 g dT
dT = 4.91 C
The final T will be 23 + 4.91 = 27.91 C
Note the use of units, you are less likely to make mistakes if you also pay attention to what units are
used.
I hoep the leads to the resistor are insulated or you will short out your 120V source ;)
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First you have to figure out at what rate power is dissipated at the resistor in the circuit by using
P = V^2 / R
Next you need to figure out how much work this resistor has done over the time interval with
W = P * t = (V^2 / R) * t
Finally you need to use the following equation to figure out how much the temperature changed by
W = m * c * delta_Temp
Where m is the mass of water (convert 1 liter of water to its mass) c is water's specific heat (it is some constant google it, make sure the units for this constant match up with the units for the mass of the water)
Now that you have delta_Temp you can find the final temperature with
Tf = To + delta_temp
Where Tf is the final temp and To is the initial temperature
P = V^2 / R
Next you need to figure out how much work this resistor has done over the time interval with
W = P * t = (V^2 / R) * t
Finally you need to use the following equation to figure out how much the temperature changed by
W = m * c * delta_Temp
Where m is the mass of water (convert 1 liter of water to its mass) c is water's specific heat (it is some constant google it, make sure the units for this constant match up with the units for the mass of the water)
Now that you have delta_Temp you can find the final temperature with
Tf = To + delta_temp
Where Tf is the final temp and To is the initial temperature
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Power dissipitated by the resistor = V^2/R = 120^2/21 = 685.714 W
Time for which present in water = 30 sec.
Therefore net energy(heat) dissipitated to water = 685.714*30 = 20571.429 J
Now Q=mc(dT) where dT is temperature change.
Therefore dT = Q/mc = 20571.429/(1000*4.186) = 4.91434 degree C (Specific heat of water is 4.186 J/(gram degree C) ; Mass of water = 1000 g assuming density to be 1g/cc.)
Hence final temperature is 23 + dT = 27.91434 degree C.
Time for which present in water = 30 sec.
Therefore net energy(heat) dissipitated to water = 685.714*30 = 20571.429 J
Now Q=mc(dT) where dT is temperature change.
Therefore dT = Q/mc = 20571.429/(1000*4.186) = 4.91434 degree C (Specific heat of water is 4.186 J/(gram degree C) ; Mass of water = 1000 g assuming density to be 1g/cc.)
Hence final temperature is 23 + dT = 27.91434 degree C.