How Does Dropping a Heavy Object into Water Affect Its Temperature?

[ryu1]

Homework Statement

1.5ton (1500kg) object is dropped from 20 meters into a bucket with 2.5L of water in it.
All the kinetic energy is transferred to the water in form of heat.
What is the temperature rise of the water?

2. how much the water temperature rises if the object lifts during it's fall, another mass(no friction involved) making it do work, spending 900.0 J before hitting the water.

The Attempt at a Solution

right not sure if i did this right but i might did:
Ek = 0.5*1500kg*(9.81ms^-2)^2 = 750kg*96.2361m/s = 72127.075m/s
72127.075m/s = 72.127KJ
specific heat of water = 4.19KJ/Kg*C
so
72.127KJ = 4.19KJ/Kg*C * 2.5Kg[=2.5L of water right?] *dT
dT=6.886C

temperature rise by 6.886 degrees Celsius.
is that actually correct?

trying the 2nd problem for now, waiting for your answer :)

right, for the 2nd problem, can we just decrease 900 from the 72127.075m/s (=72127.075J) and then calculate the whole thing exactly the same way?
arriving to dT=6.7997 Celsius?

please let me know if I am correct on this.
Thanks a lot! :)

 

[Dick]

Homework Statement

1.5ton (1500kg) object is dropped from 20 meters into a bucket with 2.5L of water in it.
All the kinetic energy is transferred to the water in form of heat.
What is the temperature rise of the water?

2. how much the water temperature rises if the object lifts during it's fall, another mass(no friction involved) making it do work, spending 900.0 J before hitting the water.

The Attempt at a Solution

right not sure if i did this right but i might did:
Ek = 0.5*1500kg*(9.81ms^-2)^2 = 750kg*96.2361m/s = 72127.075m/s
72127.075m/s = 72.127KJ
specific heat of water = 4.19KJ/Kg*C
so
72.127KJ = 4.19KJ/Kg*C * 2.5Kg[=2.5L of water right?] *dT
dT=6.886C

temperature rise by 6.886 degrees Celsius.
is that actually correct?

trying the 2nd problem for now, waiting for your answer :)

right, for the 2nd problem, can we just decrease 900 from the 72127.075m/s (=72127.075J) and then calculate the whole thing exactly the same way?
arriving to dT=6.7997 Celsius?

please let me know if I am correct on this.
Thanks a lot! :)

Try the first one again. If a is the acceleration then kinetic energy is NOT equal to (1/2)ma^2. Find a different formula.

 

[ryu1]

Try the first one again. If a is the acceleration then kinetic energy is NOT equal to (1/2)ma^2. Find a different formula.

is it Ep = 1500KG*9.81m/s^2 *20m ? then i get 294.3KJ

Is that correct? but this one is Potential energy so I got confused when talked about kinetic in the question.

 

[Dick]

is it Ep = 1500KG*9.81m/s^2 *20m ? then i get 294.3KJ

Is that correct? but this one is Potential energy so I got confused when talked about kinetic in the question.

Yes, that's the one. If the mass get stopped by hitting the water then all of the potential energy will get tranferred to kinetic energy and transferred to the water. (1/2)ma^2 isn't a kinetic energy. It's not much of anything relating to energy. The units are all wrong.

 

[ryu1]

Yes, that's the one. If the mass get stopped by hitting the water then all of the potential energy will get tranferred to kinetic energy and transferred to the water.

so I go exactly the same way from this point?
294.3KJ/10.475KJ*C = 28.095C =dT

?

it does seems to be more realistic i guess.
and the 2nd problem would be
294300-900=293.4KJ
dT = 293.4/10.475 = 28.009C
?
That is really close , but it's plausible if the change is only 900J ...is that right?
thats the last question in my homework, I have been sitting on the last 3 questions all night through (it's 7am now)
Thanks!

 

[Dick]

so I go exactly the same way from this point?
294.3KJ/10.475KJ*C = 28.095C =dT

?

it does seems to be more realistic i guess.
and the 2nd problem would be
294300-900=293.4KJ
dT = 293.4/10.475 = 28.009C
?
That is really close , but it's plausible if the change is only 900J ...is that right?
thats the last question in my homework, I have been sitting on the last 3 questions all night through (it's 7am now)
Thanks!

Sounds ok. I didn't check the exact numbers. But you dropped an awfully heavy thing a pretty large distance into a small amount of water, hence the large temperature change. For the second part you really didn't change the energy much, so the temp change should be only a little less.

 

[ryu1]

Sounds ok. I didn't check the exact numbers. But you dropped an awfully heavy thing a pretty large distance into a small amount of water, hence the large temperature change. For the second part you really didn't change the energy much, so the temp change should be only a little less.

Thank you.