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The specific heat of lead is 0.128J/(gC). If 34.2g of lead, initially at 25 degrees C,...

The specific heat of lead is 0.128J/(gC). If 34.2g of lead, initially at 25 degrees C, absorbs 1.689kJ, what will be the final temperature of the lead?

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Answer #1

q = 1.689Kj   = 1689J

m = 34.2g

c   = 0.128j/g-0c

t1   = 250C

q   = mct

q   = mc(t2-t1)

1689 = 34.2*0.128*(t-25)

1689   = 4.3776(t-25)

1689/4.3776   = t-25

385.83       = t-25

t                    = 385.83+25

t                 = 410.830C

The final temperature of lead = 410.830C

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