mikep
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A large punch bowl holds 4.75 kg of lemonade (which is essentially water) at 20.0°C. A 2.00 kg ice cube at -10.2°C is placed in the lemonade. What is the final temperature of the system, and the amount of ice (if any) remaining? Ignore any heat exchange with the bowl or the surroundings.
can someone please tell me how to solve this problem?
so far i have:
(4.75kg)(4186J/kg°C)(20°C) = 397670J
(2kg)(2090J/kg°C)(10.2°C) = 426360J
(2kg)(335000J/kg) = 670000J
can someone please tell me how to solve this problem?
so far i have:
(4.75kg)(4186J/kg°C)(20°C) = 397670J
(2kg)(2090J/kg°C)(10.2°C) = 426360J
(2kg)(335000J/kg) = 670000J