B. Extra text to get to 20 characters.
The potential energy of the spring is 6.75 J
The elastic potential energy stored in the spring is given by the equation:

where;
k is the spring constant
x is the compression/stretching of the string
In this problem, we have the spring as follows:
k = 150 N/m is the spring constant
x = 0.3 m is the compression
Substituting in the equation, we get


Therefore. the elastic potential energy stored in the spring is 6.75J .
Learn more about potential energy here:
brainly.com/question/10770261
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Answer:

Explanation:
As we know that volume is given as

so it is given in liter as

now we have six pack of such volume
so total volume is given as


so its mass is given as

now the change in temperature is given as


now the heat given to the liquid is given as



