A 2-kilogram [kg] projectile traveling at 120 meters per second [m/s] is stopped by being shot into an insulated tank containing 120 kilograms [kg] of water. If the kinetic energy of the projectile is completely converted into thermal energy with no energy lost, how much will the water increase in temperature in units of degrees Celsius [degrees Upper C]? The specific heat of water is 1 calorie per gram degree Celsius [cal/(g degrees Upper C)].

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magamezu magamezu · Answer 1 · 2019-10-09T05:31:15+02:00

Answer:

ΔT=0,028 degrees Celsius

Explanation:

There is an energy transformation, then we need to know how to find both kinds of energy, kinetic and thermal.

Kinetic energy:

[tex]E_{k} =\frac{mv^{2} }{2}[/tex]

Thermal Energy

[tex]E_{t} =cmΔT[/tex]

As the energy is always the same in a closed system, we have to equalize:

[tex]\frac{mv^{2} }{2} =cmΔT[/tex]

As we have all the information, we want to find the temperature change, then:

[tex]ΔT=\frac{m(projectile)v^{2} }{2cm(water)} \\\\ΔT=\frac{(2kg)(120 m/s)^{2} }{2(1cal/gC)(120 kg* 1000)g*(4.18 J/1 cal)}

[tex]\frac{28800 J}{1203840 J/C }[/tex]

\\\\\\\\\\ΔT=0,028 degrees Celsius[/tex]