For many purposes we can treat propane as an ideal gas at temperatures above its boiling point of . Suppose the temperature of a
sample of propane gas is lowered from to , and at the same time the pressure is decreased by . Does the volume of the sample increase, decrease, or stay the same? increase decrease stays the same If you said the volume increases or decreases, calculate the percentage change in the volume. Round your answer to the nearest percent.
1 answer:
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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
