Question

How much heat in j is giving out when 85.0g of lead cools from 200.0 c to 10.0c

Answer 1

q = - 2067.2 J of Heat is giving out when 85.0g of lead cools from 200.0 c to 10.0 c.

Step-by-step explanation:

The Specific Heat capacity of Lead is 0.128
`(J)/(g\ ^(0)C)`

This means, increase in temperature of 1 gm of lead by
`1 ^(0)\ C` will require 0.128 J of heat.

Formula Used :

`q = m.c.\Delta T`

q = amount of heat added / removed

m = mass of substance in grams = 85.0 g

c = specific heat of the substance = 0.128

`q = m.c.\Delta T` = Change in temperature

= final temperature - Initial temperature

= 10 - 200

= -
`190 ^(0)\ C`

put value in formula

q = -
`85* 0.128* 190`

On calculation,

q = - 2067.2 J

- sign indicates that the heat is released in the process