The heat is given out when 85g of lead cools from 200 °C to 10 °C the specific heat of lead is 0.11 J/g °C is 1776.5 J
The specific heat expression is given as :
Q = mc ΔT
Where,
Q = heat
m = 85 g
c = 0.11 J/g °C
ΔT = 200 °C - 10 °C
= 190 °C
Q = mc ΔT
Q = 85 × 0.11 × 190
Q = 1776.5 J
Thus, The heat is given out when 85g of lead cools from 200 °C to 10 °C the specific heat of lead is 0.11 J/g °C is 1776.5 J
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