A portion of an electric circuit connected to a 7 ohm resistor is embedded in 1.20 kg of a solid substance in a calorimeter. The external portion of the circuit is connected to a 100 volt power supply.
(a) Assuming that all the heat generated by the resistor is absorbed by the solid substance, and that it takes 4 minutes to raise the temperature from -10 to 50 degrees C, calculate the specific heat for the substance.
(b) At 50 degrees C, the substance begins to melt. The heat of fusion is 4.5 x 10^5 joules per kilogram. How long after the temperature reaches 50 degrees C will it take for all of the substance to melt?
(a) Calculate the rate of heat transfer to the substance using
P = V^2/R. Multiply by 240 seonds for the number of Joules.
That energy divided by the mass and 60 C is the specific heat
(b) time required =
(Mass)(Heat of fusion)/(Power)
We will be happy to check your work
To solve this problem, we'll need to use the equation for heat transfer formula and the equation for specific heat.
(a) First, let's calculate the heat absorbed by the solid substance. The formula for heat transfer is:
Q = mcΔT
Where:
Q = Heat energy transferred
m = Mass of the substance
c = Specific heat of the substance
ΔT = Change in temperature
In this case, we know the mass of the substance is 1.20 kg and the change in temperature is 50 degrees Celsius - (-10 degrees Celsius) = 60 degrees Celsius.
Q = (1.20 kg)(c)(60 °C)
We'll also need to calculate the total heat energy generated by the resistor, which can be found using the formula:
P = IV
Where:
P = Power generated
I = Current
V = Voltage
In this case, the voltage is 100 volts. We can assume the power generated by the resistor is dissipated as heat. So, the power is:
P = IV = (100 V)(I)
Now, because we are assuming all the heat generated by the resistor is absorbed by the solid substance, the heat transferred can be equated with the heat generated by the resistor:
Q = P × (time)
We know the time is 4 minutes, so we can write:
Q = (100 V)(I)(4 minutes)
Now, we can equate the two expressions for Q:
(1.20 kg)(c)(60 °C) = (100 V)(I)(4 minutes)
We'll assume the current is constant, so it cancels on both sides:
(1.20 kg)(c)(60 °C) = (100 V)(4 minutes)
Simplifying, we have:
(1.20 kg)(c)(60 °C) = (400 V·minutes)
Now, we need to convert minutes to seconds, since the SI unit for time is seconds. There are 60 seconds in a minute, so:
(1.20 kg)(c)(60 °C) = (400 V)(60 seconds)
Simplifying further:
(1.20 kg)(c)(60 °C) = (24000 V·s)
Now, we can solve for the specific heat (c):
c = (24000 V·s) / [(1.20 kg)(60 °C)]
(b) To find out how long it takes for all the substance to melt, we need to calculate the amount of heat required to melt it completely. Given that the heat of fusion is 4.5 × 10^5 joules per kilogram and the mass of the substance is 1.20 kg:
Heat required to melt = (1.20 kg)(4.5 × 10^5 J/kg)
Now, we know the power generated by the resistor is equal to the heat absorbed per second during the melting process:
P = (1.20 kg)(4.5 × 10^5 J/kg) / time
Given that the voltage is 100 volts:
100 V(I) = (1.20 kg)(4.5 × 10^5 J/kg) / time
Now, solving for time:
time = (1.20 kg)(4.5 × 10^5 J/kg) / (100 V(I))
This equation will give us the time required for all the substance to melt at 50 degrees Celsius.