Question 1085178
A) Let x be the final temperature of the mixture
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The iron is warmer, so its delta temperature is 40 - x
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The water's delta temperature is x - 20 
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the energy amount going out of the iron is equal to the energy amount going into the cool water. This means,
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q(lost) = q(gain)
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we are given that the mass(m) of the iron is the same as the water
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we know
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q = m * delta temperature * specific heat
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m * (40 - x) * 0.113 = m * (x - 20) * 1
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now we solve for x
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4.52 - 0.113x = x - 20
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-1.113x = -24.52
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x = 8.1222
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final temperature is 8.1222 C
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B) Our balance equation, q(lost) = q(gain) is
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m * (10 - x) * 1 = m * (x - 0) * 1
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10 - x = x
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2x = 10
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x = 5
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we expect the final temperature to be 5 C
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