A large punch bowl holds 3.95 kg of lemonade (which is essentially water at 20 d

Question

A large punch bowl holds 3.95 kg of lemonade (which is essentially water at 20 degrees Celsius. A 0.055kg ice cube at -3 degrees Celsius is placed in the lemonade. What are the final temperature of the system, and the amount of ice (if any) remaining? Ignore any heat exchange with the bowl or the surroundings. You decide that this temperature isn’t good enough, and you want there to still be 20 g of ice in the punch bowl in equilibrium (which means the equilibrium temperature of the punch/ice is 0 degree Celsius. You decide to add more ice! How much ice would you need to add to the initial 20 degree Celsius lemonade in order to achieve the desired final state? A friend watches your party calculations and says that your lemonade will obviously be too watery. He declares that you should instead use the same original mass (0.055kg) of ice, but instead use colder ice. This, at first, seems reasonable, but then you realize that your friend is mistaken. Why is your desired final equilibrium state impossible using colder ice?

Explanation / Answer

The amount of heat removed from the lemonade to bring it down to -3 oC

Q = mcT

Q = (3.95)(4186)(20-(-3))

=3.8 * 10^5 J

Q_ice = m_ice L_f = 0.055 ( 33.5 * 10^4) = 1.84 * 10^4 J

Q_water is greater than Q_ice so all the ice in hte bowl will melt

Q_net = Q_water-Q_ice = 3.8 * 10^5 - 0.184 * 10^5 = 3.616 * 10 ^5

Q_net is then the heat to warm up the water in the bowl and the water just melted

Q_net = ( m_water + m_ice) c _ water ( Tf- (-3)

3.616 * 10^5 = ( 3.95 + 0.055) ( 4186) ( T_f + 3)

T_f = 18.56 C

Q = 313113 J

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