Answer:
q = - 2067.2 J of Heat is giving out when 85.0g of lead cools from 200.0 c to 10.0 c.
Explanation:
The Specific Heat capacity of Lead is 0.128
This means, increase in temperature of 1 gm of lead by will require 0.128 J of heat.
Formula Used :
q = amount of heat added / removed
m = mass of substance in grams = 85.0 g
c = specific heat of the substance = 0.128
= Change in temperature
= final temperature - Initial temperature
= 10 - 200
= -
put value in formula
q = -
On calculation,
q = - 2067.2 J
- sign indicates that the heat is released in the process