4. Consider an aluminum calorimeter (m = 0.50 kg) with 4.00 kg of water. Both th

Question

4. Consider an aluminum calorimeter (m = 0.50 kg) with 4.00 kg of water. Both the water and the calorimeter begin at a temperature of 45.0 C. An unknown amount of ice (@0 C ) is added to the calorimeter until the temperature reaches 5.00 C

a. How much heat does the original water in the calorimeter gain as it cools to 5.00 C?

b. How much heat does the ice absorb as it melts and warms to 5.00 C?

c. How much heat does the aluminum cup give up as it cools to 5.00 C?

d. How much ice is used to cool the calorimeter and water to a final temperature of 5.00 C?

e. How much steam at 125.0 C would you need to bring this system back to 45.0 C after it has been cooled by the ice?

Explanation / Answer

a)

heat loss by original water = mC(dT) = 4000*1*40 = 160 kcal

b)

heat gain by ice = heat loss by water anf calorimeter

= 0.215*500*40 + 1*4000*40

  = 164.3 kcal

c)

heat given by aluminium cup = 0.215*500*40 = 4.3 kcal

d)

let mass of ice be x kg

latent heat of melting + dT from 0 to 5 = heat gain by ice

x*1000*79.8 + x*1000*1 = 164300 cal

x = 2.0334 kg

e)

heat required to bring the systme back to 45 C = 6033.4*1*40 + 0.215*500*40

= 245.636 cal

let mass of steam require = y

thus

y*0.48*25 + y*541.68 + y*1*55 = 245.636

y = 0.40358 kg steam

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