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thermodynamics question please help

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
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?
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
Reply 3
Avatar for lfc_all_day
lfc_all_day
OP
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?
(edited 9 years ago)
Reply 4
Avatar for lfc_all_day
lfc_all_day
OP
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 ??
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.
how much you work out the entropy change of water for this question ??
what would the entropy change of the iron ball be ?

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