1)
heat required to take ice from -17 oC to 0 oC,
Q1 = Mi*Ci*delta Ti
= 65*1.06*(17)
= 1171.3 J
Heat required to take water from 29.2 oC to 0 oC,
Q2 = Mw*Cw*delta Tw
= 256*4.184*29.2
= 31276 J
heat required to melt whole ice,
Q3 = Mi*Li
= 65*333
= 21645 J
Q1+Q3 = 1171.3 + 21645 = 22816.3 J < Q2
So whole of ice will melt
let final temperature be ToC
Q1 + Q3 + heat required by melted ice = heat released by
water
22816.3 + 65*4.184*T = 256*4.184*(29.2-T)
22816.3 + 267.8*T = 31276 - 1071.1*T
T = 6.3 oC
Answer: 6.3 oC
I am allowed to answer only 1 question at a time
Get Answers For Free
Most questions answered within 1 hours.