A 180 g copper bowl contains 255 g of water, both at 20.0°C. A very hot 300 g co

Question

A 180 g copper bowl contains 255 g of water, both at 20.0°C. A very hot 300 g copper cylinder is dropped into the water, causing the water to boil, with 5.95 g being converted to steam. The final temperature of the system is 100°C. Neglect energy transfers with the environment. PLEASE READ THIS NOTE I HAVE FOUND THE SOLUTIONS FOR THIS ALL OVER THE WEB BUT FOR SOME REASON WHEN I ENTER IT IN TO MY CALCULATOR IT IS INCORRECT IF ANYONE OUT THEIR CAN ACTUALLY EXPLAIN THE PROBLEM RATHER THAN COPY IT WOULD BE MUCH APPRECIATED.

What is the original temperature of the cylinder °C ?

Explanation / Answer

Q(lost) = Q(gain) m x s x ?t = 468892.80 + 5236 300 x 10^-3 x 385 x (t - 100) = 474128.8 t = 4205.01 C

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