Asked by Mark
A cube of ice is take from the freezer at -8.5C and placed in a 95g calorimeter aluminum cup containing 310g of water at 20degC. The final situation is observed to be all water at 17degC. What was the mass of the ice cube?
Answers
Answered by
bobpursley
The sum of heats gained is zero.
HeatgainedbyicewarmigtoOC + heatgainedmeltingIcd+ heat gained by aluminum+heat gained by original water=0
massice*Cice*(0-8.5C)+massice*Hf+95*Calum*(17-20)+310*Cwater*(17-20)=0
solve for mass of ice.
HeatgainedbyicewarmigtoOC + heatgainedmeltingIcd+ heat gained by aluminum+heat gained by original water=0
massice*Cice*(0-8.5C)+massice*Hf+95*Calum*(17-20)+310*Cwater*(17-20)=0
solve for mass of ice.
Answered by
Sherileen Eva
19.276g
Answered by
Anonymous
m cice
0 − (−8.5)
| {z }
ice from −8.5
◦C to 0◦C
+ m Lwater
| {z }
ice melts at 0◦C
+ m cwater
17 − 0
| {z }
water from 0◦C to 17◦C
= 0.300cwater
20 − 17
| {z }
water from 20◦C to 17◦C
+ 0.075cAl
20 − 17
| {z }
Al from 20◦C to 17◦C
⇒ m
2090
8.5 + m
3.33 × 105
+ m
4186
17 = 0.300
4186
3 + 0.075
900
3
⇒ m
421,927
= 3969.9
0 − (−8.5)
| {z }
ice from −8.5
◦C to 0◦C
+ m Lwater
| {z }
ice melts at 0◦C
+ m cwater
17 − 0
| {z }
water from 0◦C to 17◦C
= 0.300cwater
20 − 17
| {z }
water from 20◦C to 17◦C
+ 0.075cAl
20 − 17
| {z }
Al from 20◦C to 17◦C
⇒ m
2090
8.5 + m
3.33 × 105
+ m
4186
17 = 0.300
4186
3 + 0.075
900
3
⇒ m
421,927
= 3969.9
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