Respuesta :
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 [tex]\frac{J}{g\ ^{0}C}[/tex]
This means, increase in temperature of 1 gm of lead by [tex]1 ^{0}\ C[/tex] will require 0.128 J of heat.
Formula Used :
[tex]q = m.c.\Delta T[/tex]
q = amount of heat added / removed
m = mass of substance in grams = 85.0 g
c = specific heat of the substance = 0.128
[tex]q = m.c.\Delta T[/tex] = Change in temperature
= final temperature - Initial temperature
= 10 - 200
= -[tex]190 ^{0}\ C[/tex]
put value in formula
q = - [tex]85\times 0.128\times 190[/tex]
On calculation,
q = - 2067.2 J
- sign indicates that the heat is released in the process
