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Q: If 1600 cal of heat are added to 50 g of water initially at a temperature of 10 celsius what is the final temperature of the water?
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What is the final temperature if 200 Joule of heat is removed from 50N block of ice initially at 25 degrees Celsius?

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How 750 calories of heat are added to 250 grams of Lead c equals 0.031 cal g C initially at a temperature of 28.0 C What will be the final temperature of this piece of Lead?

I need to make some conversions for my own convenience. 750 calories (4.184 Joules/1 calorie) = 3138 Joules ----------------------- Lead specific heat (c) = 0.160 J/gC Now, q = mass * specific heat * change in temperature 3138 Joules = (250 g Pb)(0.160 J/gC)(Tf - 28.0 C) 3138 J = 40Tf- 1120 4258 = 40Tf 106 Celsius = Final temperature of lead ----------------------------------------------------