Asked by aj

Determine the resulting temperature when 150g of ice at 0°C is mixed with 300g of water at 50°C
Answered by drwls
Heat gained by ice (including after melting) = Heat lost by original liquid water.

Solve for final temperature, T

150(80 + T) = 300(50 - T)
12,000 + 150 T = 15,000 -300 T
450 T = 3000
T = 6.7 C
Answered by ridhi
thanks :)
Answered by ESTHER
6.7
Answered by Anonymous
GUINEA
Answered by Sansi baisla
Can you please secribe the value 80+T is what and from where it come
Answered by Anonymous
From Where Do We Have 80+T
Answered by Spending
Where do you get 80-T from?
Answered by Jophanie
(150g)(80cal/g°c)= -(300g)(cal/g°c)(T-50°c)

1200cal= -300cal T + 15000 cal
300cal T = 15000 - 12000

300cal T= 3000cal
T=3000cal/300cal °c
Answered by Raphael Orawai
Where Did You Get 80 + T ?
Answered by Adeosun opeyemi
latent energy - needed to melt ice, is equivalent to raising the temperature of water by 80°C.
While melting, 150g of ice will therefore lower the temperature of twice as much by half as much, or 40°C
The question is therefore equivalent to mixing 150g of water at 0°C with 300g of water at 10°C.
6,7°C
Answered by Precious
Please where did the 80 come from? Am confused
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