# thermal physics help

Watch
Announcements

Page 1 of 1

Go to first unread

Skip to page:

An ice cube of mass 0.010 kg at a temperature of 0 °C is dropped into a cup containing 0.10 kg of water at a temperature of 15 °C.

What is the maximum estimated change in temperature of the contents of the cup?

specific heat capacity of water = 4200 J kg−1 K−1

specific latent heat of fusion of ice = 3.4 × 105 J kg−1

I'm not really sure how to do this question, I keep getting different answers each time.

for ice I did: ml+mcdeltatheta=E

for water: E=mcdeltatheta

I tried to equate both of them but I'm getting different values, please help

What is the maximum estimated change in temperature of the contents of the cup?

specific heat capacity of water = 4200 J kg−1 K−1

specific latent heat of fusion of ice = 3.4 × 105 J kg−1

I'm not really sure how to do this question, I keep getting different answers each time.

for ice I did: ml+mcdeltatheta=E

for water: E=mcdeltatheta

I tried to equate both of them but I'm getting different values, please help

0

reply

Report

#2

Meh i got a reasonable answer.

So first you have to reduct the potential energy of the ice from the water. So m(ice)L = m(water)cΔt and substituted to calcualte the temperature change

That will give a temperature change of about -8 degrees.

Next you have water .01kg at 273 kelvin and .1 kg at 280 kelvin. Calculate there respective energies using E = mct for both sources. Then sum the energies and divide by specific heat capcity and the total mass. You should get about 279 K so the temperature of the cup has dropped 9 degrees.

So first you have to reduct the potential energy of the ice from the water. So m(ice)L = m(water)cΔt and substituted to calcualte the temperature change

That will give a temperature change of about -8 degrees.

Next you have water .01kg at 273 kelvin and .1 kg at 280 kelvin. Calculate there respective energies using E = mct for both sources. Then sum the energies and divide by specific heat capcity and the total mass. You should get about 279 K so the temperature of the cup has dropped 9 degrees.

0

reply

(Original post by

Meh i got a reasonable answer.

So first you have to reduct the potential energy of the ice from the water. So m(ice)L = m(water)cΔt and substituted to calcualte the temperature change

That will give a temperature change of about -8 degrees.

Next you have water .01kg at 273 kelvin and .1 kg at 280 kelvin. Calculate there respective energies using E = mct for both sources. Then sum the energies and divide by specific heat capcity and the total mass. You should get about 279 K so the temperature of the cup has dropped 9 degrees.

**Dominininc**)Meh i got a reasonable answer.

So first you have to reduct the potential energy of the ice from the water. So m(ice)L = m(water)cΔt and substituted to calcualte the temperature change

That will give a temperature change of about -8 degrees.

Next you have water .01kg at 273 kelvin and .1 kg at 280 kelvin. Calculate there respective energies using E = mct for both sources. Then sum the energies and divide by specific heat capcity and the total mass. You should get about 279 K so the temperature of the cup has dropped 9 degrees.

0

reply

quick question for Ice don't you do ml+Mcdeltatheta since its turning from solid to liquid? and how did you get .01kg of water to be 273 Kelvin and same as ice? I'm not really too sure to be honest

Last edited by dwrfwrw; 2 weeks ago

0

reply

Report

#5

Oh my i just realised i did the question wrong

Basically calculate the total energy of the system. so E= .01c x 273 + .1l +.1c x 288

The total energy in the system is like 166426 J i believe. Take away the amount of energy that will go into into internal energy so 166426- .11 x L = 129026J This is the amount of energy that goes into the kinetic store so it can be subsititued into the equation. 129026/.11c = 279.27

so the temperature change is actually 288-279.27 = 8.73

Basically calculate the total energy of the system. so E= .01c x 273 + .1l +.1c x 288

The total energy in the system is like 166426 J i believe. Take away the amount of energy that will go into into internal energy so 166426- .11 x L = 129026J This is the amount of energy that goes into the kinetic store so it can be subsititued into the equation. 129026/.11c = 279.27

so the temperature change is actually 288-279.27 = 8.73

Last edited by Dominininc; 2 weeks ago

0

reply

(Original post by

Oh my i just realised i did the question wrong

Basically calculate the total energy of the system. so E= .01c x 273 + .1l +.1c x 288

The total energy in the system is like 166426 J i believe. Take away the amount of energy that will go into into internal energy so 166426- .11 x L = 129026J This is the amount of energy that goes into the kinetic store so it can be subsititued into the equation. 129026/.11c = 279.27

so the temperature change is actually 288-279.27 = 8.73

**Dominininc**)Oh my i just realised i did the question wrong

Basically calculate the total energy of the system. so E= .01c x 273 + .1l +.1c x 288

The total energy in the system is like 166426 J i believe. Take away the amount of energy that will go into into internal energy so 166426- .11 x L = 129026J This is the amount of energy that goes into the kinetic store so it can be subsititued into the equation. 129026/.11c = 279.27

so the temperature change is actually 288-279.27 = 8.73

0

reply

Report

#7

15 degrees is 288 kelvin, the standard unit of temperature. with -273 being 0 kelvin. Ice is 273 because thats 0 degrees c in kelvin

Ice and water can have the same temperature. If they have the same kinetic energy, however water has more potential energy. But this doesnt effect what appears on a themometer.

Ice and water can have the same temperature. If they have the same kinetic energy, however water has more potential energy. But this doesnt effect what appears on a themometer.

0

reply

X

Page 1 of 1

Go to first unread

Skip to page:

### Quick Reply

Back

to top

to top