Recent content by twilos

Homework Statement A stack of identical coins is all at a temperature of 100 C. The coins are transferred to a container of room temperature water. After thermal equilibrium is reached the final temperature is 35 C. The Heat lost by a single coin is less/greater/equal to the heat lost by all...

so i am right!? awesome it does weigh more because the water displaced is 8N worth also right?

Homework Statement Before you submerge a block (25N) in water, you place the container of water on a scale. The scale reads 100N. You then immerse the block in the water, the block is also hung on a scale that reads 17N. Is the scale reading greater than, less than, or equal to 100N? (...

(2.5 kg)(910 J/kg K)(T - (-135C)) = - (0.5 kg)(4190 J/kg K)( T - 25C) +(0.5 kg)(334000 J/kg) + (2100 J/ kg K)( T - 0C) i believe all the numbers are correct however I am always confused because of the positives and negatives... and the left side is gaining heat and the right side is losing...

damn i can't get it to work out like i am really bad with the algebra lol anyone help me simplify this please?

okay so the following i did: Qal = (2.5kg)(910 J/kg K) (135C) = 307125 J Qw = (0.5kg)(4190 J/kg K) (25C) = 52375 J Therefore the amount of heat for water is less than Aluminum so all of the water turns into ice? then i can just do : (2.5 kg)(910 J/kg K)(T - (-135C)) = - (0.5 kg)(4190 J/kg...

that is necessary to find the temperature final? And if its partially converted doing what you said will give a ratio of how much ice is in the water?

(2.5 kg)(910 J/kg K)(T - (-135C)) = - (0.5 kg)(4190 J/kg K)( 0 - 25C) +(x)(334000 J/kg) + (2100 J/ kg K)( T - 0C) Is this what your talking about? Because then there's 2 variables X because you don't know the amount of water that changes into ice...?

Homework Statement Several small piece of aluminum, having total (combined) mass of 2.5 kg are placed in liquid nitrogen and, when removed, are at a temperature of -135C. The aluminum is transferred to an insulating container with 0.50 kg of water. The water is at 25.0C [ Ignore heat lost from...

Awesome i actually understood the concept now... =) so we're splitting Part A into the Before and Part B into the After. And Part B will be looked at again with the conservation of energy as "the KE going back into Spring PE and gravity PE and the expense of friction." correct?

Just to clarify, that was all for part A right? Now to find the max compression of the spring do i keep all the variables still i.e kinetic, potential gravitational, potential spring, friction? And do i solve for k or x/D?

for Part B where you find the maximum compress I believe i am wrong and your actually finding X which is the distance the spring is compressed. Therefore when you find Vf you substitute it back into the conservation of energy equation and solve for X or D which i used both. And there will...

Homework Statement A 2.1 kg is released from rest from the bottom of a 45 degree inclined ramp. The package is attached to an ideal spring K = 35 N/m that is attached to the top of the incline causing the package to be launched up the incline. The coefficients of friction between the package...