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D www.google.com k Announcements meta x Assignments x physics guestion LCheg x C Physics Mastering Physics: PS-1-Preparation Google Chrome Secure https:// session.masteringphysics.com /myct/itemView? assignment ProblemID 37417164 PHYS1052 General Physics II PS-1-Praparation PS-1-Prep-1-SS2013 Mastering Phys O My Courses PS-1-Prep-1-SS2013 Course Home Part A A 145 g aluminum cup contains 145 g of water and If you add 145 g of hot water at 90 C what is the final temperature at thermal equilibrium? Syllabus both are at 10 C. Express your answer using three significant figures. Assignments Scores Course Tools Final Temperature 48.9 eText Submit My Answers Give Up Study Area User Settings incorrect Try Again, 5 attempts remaining Part B How much hot water at 90 C needs to be added to achieve a final temperature of 50 C? Mass Hot Water 176 g Submit My Answers Give Up All attempts used, correct answer displayed Ask me anything blac Help I Close previous l 1 of 4 l next CTIONS nt View nt Mew nt View 1034 PM 9/2017

Explanation / Answer

here,

mass of alumunium , ma = 0.145 kg

mass of water , m1 = 0.145 kg

initial temprature , T1 = 10 degree C

A)

mass of water , m2 = 0.145 kg

T2 = 90 degree C

let the final temprature be Tf

heat lost by hot water = heat gained by cold water + heat gained by cup

m2 * Cw * ( T2 - Tf) = m1 * Cw * ( Tf - T1) + ma * Ca * ( Tf - T1)

0.145 * 4186 * ( 90 - Tf) = 0.145 * 4186 * (Tf - 10) + 0.145 *910* ( Tf - 10)

solving for Tf

Tf = 46.08 degree C

the final temprature is 46.08 degree C

b)

final temprature , Tf = 60 degree C

let the mass of hot water be m2

heat lost by hot water = heat gained by cold water + heat gained by cup

m2 * Cw * ( T2 - Tf) = m1 * Cw * ( Tf - T1) + ma * Ca * ( Tf - T1)

m2 * 4186 * ( 90 - 60) = 0.145 * 4186 * (60 - 10) + 0.145 *910* ( 60 - 10)

solving for m2

m2 = 0.29 kg

the mass of hot water is 0.29 kg or 294.2 g

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