A coffe cup calorimeter contains 480 grams of water at 25 C 380 grams of water a

Question

A coffe cup calorimeter contains 480 grams of water at 25 C 380 grams of water at 53.5 C is added 525 grams of water at 65.5 C is added Assuming the heat absorbed by the styrofoam is negligible calculate the expected final temperature The specific heat of water is 4.184 J/g*C A coffe cup calorimeter contains 480 grams of water at 25 C 380 grams of water at 53.5 C is added 525 grams of water at 65.5 C is added Assuming the heat absorbed by the styrofoam is negligible calculate the expected final temperature The specific heat of water is 4.184 J/g*C 380 grams of water at 53.5 C is added 525 grams of water at 65.5 C is added Assuming the heat absorbed by the styrofoam is negligible calculate the expected final temperature The specific heat of water is 4.184 J/g*C

Explanation / Answer

step-1

heat lost by hot water = gained by cold water

    m*s*DT = m*s*DT

380*4.184*(53.5-T) = 480*4.184*(T-25)

   T = final temperature of mixture = 37.59 c

step -2

   heat lost by hot water = gained by cold water

    m*s*DT = m*s*DT

   525*4.184*(65.5-T) = (480+380)*4.184*(T-37.59)

Tf = final temperature of mixture = 48.17 C

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