6. A certain calorimeter is made of 100 g of copper (c-390 J/kg·°C) and contains

Question

6. A certain calorimeter is made of 100 g of copper (c-390 J/kg·°C) and contains 0.180 kg of water (c = 4186 J/kg·°C) and 20.0 g of ice (L,-3.33 × 105 J/kg) in thermal equilibrium and at atmospheric pressure. A 0.800 kg block of lead (c-130 J/kg·°C) with a temperature of 275°C is now dropped into the calorimeter. Assuming that the calorimeter is perfectly insulated so that no energy is exchanged by heat with the surroundings, a. b. What is the final temperature of the mixture in the calorimeter? Which is more useful for storing thermal energy, the block of lead or the water in the calorimeter? [Answers: (a) 22.4°C (b) you should come up with the answer to this ]

Explanation / Answer

-Qhot = Qcold

-Qlead = Qcopper + Qwater+ Qice

thermal energy required to melt ice:

Qice = m*LH = 20*333 = 6660 J

Qlead = m*C*(Tf-Tl) = 800*0.13 * (Tf-275) -  20*333

now..

Qhot = Qcold

total mass of water = melted ice + water = 20 + 180 = 200

-800*0.13 * (Tf-275) -  20*333 = Qwater + Qcopper

-800*0.13 * (Tf-275) -  20*333 = m*C*(Tf-Tw)+ mc*Cc*(Tf-Tw)

-800*0.13 * (Tf-275) -  20*333 = 200*4.184*(Tf-0)+ 100*0.39*(Tf-0)

-800*0.13 * (Tf-275) -  20*333 = (200*4.184 + 100*0.39)Tf

-104Tf + 275*104 - 6660 = 875.8Tf

Tf(104 + 875.8) =  275*104 - 6660

Tf = ( 275*104 - 6660 ) / (104 + 875.8) = 22.40°C

b)

water is much more efficient, since it reuqire smore heat per un it °C and mass

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