Get expert insights and community support for your questions on IDNLearn.com. Receive prompt and accurate responses to your questions from our community of knowledgeable professionals ready to assist you at any time.
Answer:
Explanation:
From the given information:
It is pertinent to understand that few things need to be considered in this process. The huge metallic vessel needs warming alongside legitimate guidelines of temperature for powerful heat need of the container(organism) in the first place. Simultaneously it is additionally appropriate to chill off the substance of the vessel and justified the cooling limit. Kindly note that it is temperature is additionally related to the encompassing pressing factor and diminished temperature is a characteristic of lower pressure, which can be considered surrounding pressure (one or lower than 1 air pressure).
SO;
Specific gravity = 1
mass of water = 135 kg which is equivalent to 135 liters
To [tex]cm^2[/tex]; we have:
1 liter = 100 [tex]cm^2[/tex]
5 liter = 292 [tex]cm^2[/tex]
∴
135 L will yield 2631.616 [tex]cm^2[/tex]
However, converting this into MegaJoules
Recall that:
cal/[tex]cm^2[/tex] = 1 calorie or 4.19 Joules
4.19 Joules = 4.19 × 10⁻⁶ MegaJoules
Hence;
2631.616 [tex]cm^2[/tex] = (2631.616 × 4.19 × 10⁻⁶) Mega Joules
= 0.0110264878
= 11026.47 × 10⁻⁶ Mega Joules per day