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Degree: Physics With a years Professional Experience (Masters) - Exeter University
FULL QUESTION: Two balls are attached to the bottom two separate rods with a type of glue that melts once it reaches 30°C. One rod is made of 1Kg of Iron and the other of 3.5Kg of lead. Given that the energy of both rods is increase by 100J/s simultaneously, and they both start at 20°C.
(The Specific Heat Capacities are: Iron 450J / kg °C and Lead 130J / kg °C)
a) Which ball drops first.
b) The time taken until both balls have dropped.
For part a) the first thing you must do is calculate the energy required to increase each rod by one degree Celsius. To do this you use the equation:
E = m × c × θ
And as θ=1°C
E = m × c
So for the Iron rod m=1Kg and c=450J/Kg°C, Therefore:
E = 450 × 1 = 450J
And for the Lead rod m=3.5Kg and c=130J/Kg°C, Therefore:
E = 130 × 3.5 = 455J
This means that the Iron rod takes less energy per degree to heat up, and, because each rod receives the same energy per second, the Iron rod will reach 30°C and therefore that ball will be the first to fall.
For part b), now you know that the ball attached to the iron rod drops first and thus the ball attached to the lead drops second, the time taken for both balls to drop will be the time taken for the ball attached to the lead rod to drop.
First you need to find the energy required to heat the lead rod to 30°C.
So we know that the rod needs to be at 30°C and starts at 20°C, therefore θ= 30 - 20 = 10°C.
Now the variable for the lead are m=3.5 Kg, c=130 J/Kg°C and θ=20°C.
Input these into the Specific heat capacity equation you get:
E = 3.5 × 130 × 10 = 4550J
This is the required energy for the lead rod to increase 10°C.
The question states that the rods receive 100J per second; therefore, all that is needed is to divide the required energy by the energy per second:
This gives the seconds required for the ball attached to the Lead rod to drop, which is also the time taken for both balls to drop and is the answer to part b.see more