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A person tries to heat up her bath water by adding 5.0 L of water at 80 degrees celsius  to 60 L of water at 30 degrees celsius. What is the final temperature of the water?
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Added Mon, 03 Aug '15
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let T is the final temperature

heat lost = heat gained

m1*C*(Ti-Tf) = m2*C*(Tf-Ti)

5*4186*(80-T) = 60*4186*(T-30)

80- T = 12*T - 360

13*T = 360+80

T = 440/13

= 33.846 degrees celsius
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Added Mon, 03 Aug '15
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