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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?
Wiki User
∙ 13y ago
I need to make some conversions for my own convenience.
750 calories (4.184 Joules/1 calorie)
= 3138 Joules
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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
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Wiki User
∙ 13y ago
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