1) A large punch bowl holds 3.10 kg of lemonade (which is essentially water) at
ID: 2014685 • Letter: 1
Question
1) A large punch bowl holds 3.10 kg of lemonade (which is essentially water) at 21.0 degs C. A 6.00×10-2 kg ice cube at -13.0 degs C is placed in the lemonade. What is the final temperature of the system? Ignore any heat exchange with the bowl or the surroundings.____=degrees C
2)An 800-g iron block is heated to 390 degrees C and placed in an insulated container (of negligible heat capacity) containing 35.0 g of water at 25.0 degrees C. What is the equilibrium temperature of this system? The average specific heat of iron over this temperature range is 560 J/(kg*K) .
____=degrees C
Explanation / Answer
To do both these questions, it's important to know three thermodynamics principles:
1st: given no heat exchange between the system (bowl, insulated container) and the surroundings (everything else), ALL the heat LOST by one substance in the container is GAINED by the other substance. Heat doesn't just disappear, so if it's lost by the lemonade it must be gained by the ice (or block transfers heat to water, etc.)
2nd: the formula for heat gain/loss is "Q = m*c*(change in temp) and you can set the heat lost by one item equal to the heat gained by another.
3rd: "c" is the specific heat, aka the amount of heat a substance can absorb into itself before its temperature changes. "mc" is the heat capacity of a given mass of a specific substance. For water this is:
1 calorie/(gram*degrees celsius) OR 4.184 Joules/(gram*degrees celsius)
QUESTION 1:
**this answer assumes your teacher doesn't want you to take into account the melting of the ice!
Q lost = Q gained
(mc/T) lost by lemonade = (mc/T) gained by ice =
(3100 grams)(4.184)(21 - X) = (60.0 grams)(4.184)(X - (-13))
X is the final temp...
(Hint: it's between the two temps but not precisely equal)
QUESTION 2:
(800)(.56)(390-X) = (35)(4.184)(X-25)
X is again the final temp and everything has been converted into grams
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