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#1
A 5g iron ball at 95°C is added to a well-insulated beaker containing 10g of water and allowed to thermally equilibrate, reaching a temperature of 45°C.

Cp(Iron)=25.1JK-1mol-1
Cp(water)=75.3JK-1mol-1

What was the original temperature of the water?

i assume i have to use the q=m*c*deltaT equation but i dont know how to use it
0
4 years ago
#2
(Original post by lfc_all_day)
A 5g iron ball at 95°C is added to a well-insulated beaker containing 10g of water and allowed to thermally equilibrate, reaching a temperature of 45°C.

Cp(Iron)=25.1JK-1mol-1
Cp(water)=75.3JK-1mol-1

What was the original temperature of the water?

i assume i have to use the q=m*c*deltaT equation but i dont know how to use it
Use it twice. Final temperature = t
Once for the iron cooling down (heat lost = mc(95-45))
Once for heat gained by the water.(heat gained = mc(45-t))

Do you know how to get t from this?
1
4 years ago
#3
(Original post by lfc_all_day)
A 5g iron ball at 95°C is added to a well-insulated beaker containing 10g of water and allowed to thermally equilibrate, reaching a temperature of 45°C.

Cp(Iron)=25.1JK-1mol-1
Cp(water)=75.3JK-1mol-1

What was the original temperature of the water?

i assume i have to use the q=m*c*deltaT equation but i dont know how to use it
Energy is conserved, so using E=mcT, Initial energy of the iron ball + Initial energy of the water = Final energy of ball + Final energy of water

Rearrange this and you can work out energy of the water before the ball is added, and then apply E=MC*Delta T
0
#4
(Original post by Stonebridge)
Use it twice. Final temperature = t
Once for the iron cooling down (heat lost = mc(95-45))
Once for heat gained by the water.(heat gained = mc(45-t))

Do you know how to get t from this?
hi, thanks for iron i got q = 6275, then for water i got 33885-753t, i know i have to now work out what t is but how do i do that from here?
0
#5
(Original post by lfc_all_day)
hi, thanks for iron i got q = 6275, then for water i got 33885-753t, i know i have to now work out what t is but how do i do that from here?
im unsure if this is correct but would i have to do (33885-6275)/753 ??
0
4 years ago
#6
(Original post by lfc_all_day)
im unsure if this is correct but would i have to do (33885-6275)/753 ??
Yes. Assuming the numerical values are correct, that's the correct method.
1
4 years ago
#7
how much you work out the entropy change of water for this question ??
0
2 years ago
#8
what would the entropy change of the iron ball be ?
0
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