Please answer BOTH PARTS IN THE \"EXERCISE\" SECTION. Thank you in advance! i sa

Question

Please answer BOTH PARTS IN THE "EXERCISE" SECTION. Thank you in advance!

i safari File Edit View History Bookmarks Window Help Q 60% i -E Tue 3:36:16 PM webassign.net Phy 1010 Ch 11 Hom-W18 Chegg Study Guided Solutions and Study Help All the ice will melt. Anki O The final temperature will be lower. Twice as much of the ice will melt. Half as much of the ice will melt. The final temperature will be the same. O The final temperature will be higher. Screen Shot 018-03..6.13 PM PRACTICE IT Use the worked example above to help you solve this problem. A 5.40 kg block of ice at 0°C is added to an insulated container partially filled with 12.0 kg of water at 15.0°C. (a) Find theal temperature, neglecting the heat capacity of the container. eC Screen Shot 2018-03...6.06 PM (b) Find the mass of the ice that was melted 3.99 Your response differs from the correct answer by more than 10%. Double check your calculations. kg EXERCISE HINTS: GETTING STARTED IM STUCK! Ir 8.10 kg of ice at -5.00°C is added to 12.0 kg of water at 15°C, compute the final temperature. ec How much ice remains, if any? kg Need Help? Talk to a Tutor 27

Explanation / Answer

Practice it.

b) Heat gained by Ice = Heat lost by water

m_Ice*Lf = m_water*C_water*(15 - 0 )

m_Ice = m_water*C_water*(15 - 0 )/Lf

= 12*4186*15/(3.33*10^5)

= 2.26 kg

Excercise :

a)T = 0 degrees celsius

b) Heat gained by Ice = Heat lost by water

m_Ice*C_Ice*(0 - (-5)) + m_Ice*Lf = m_water*C_water*(15 - 0 )

m_Ice = (m_water*C_water*(15 - 0 )/(C_Ice*5 + Lf)

= 12*4186*15/(2100*5 + 3.33*10^5)

= 2.19 kg

mass of the Ice remaining = 8.1 - 2.19

= 5.91 kg

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