Question

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

Answer 1

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 $\frac{J}{g\ ^{0}C}$

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 = -  $85\times 0.128\times 190$

On calculation,

q = - 2067.2 J

- sign indicates that the heat is released in the process