A group of students decides to set up an experiment in which they will measure the specific heat of a small amount of metal. The
1 answer:
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Answer:
240 ohms
Explanation:
From Ohms law we deduce that V=IR and making R the subject of the formula then R=V/I where R is resistance, I is current and V is coltage across. Substituting 120 V for V and 0.5 A for A then
R=120/0.5=240 Ohms
Alternatively, resistance is equal to voltage squared divided by watts hence
Answer:
W = 3.12 J
Explanation:
Given the volume is 1.50*10^-3 m^3 and the coefficient of volume for aluminum is β = 69*10^-6 (°C)^-1. The temperature rises from 22°C to 320°C. The difference in temperature is 320 - 22 = 298°C, so ΔT = 298°C. To reiterate our known values we have:
β = 69*10^-6 (°C)^-1 V = 1.50*10^-3 m^3 ΔT = 298°C
So we can plug into the thermal expansion equation to find ΔV which is how much the volume expanded (I'll use d instead of Δ because of format):
So ΔV = 3.0843*10^-5 m^3
Now we have ΔV, next we have to solve for the work done by thermal expansion. The air pressure is 1.01 * 10^5 Pa
To get work, multiply the air pressure and the volume change.
W = 3.12 J
Hope this helps!