Question

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​)].

Answer 1

ΔT=0,028 degrees Celsius

Step-by-step explanation:

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

Kinetic energy:

`E_(k) =(mv^(2) )/(2)`

Thermal Energy

`E_(t) =cmΔT`

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

`(mv^(2) )/(2) =cmΔT`

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

`ΔT=(m(projectile)v^(2) )/(2cm(water)) \\\\ΔT=((2kg)(120 m/s)^(2) )/(2(1cal/gC)(120 kg* 1000)g*(4.18 J/1 cal)) </p><p></p><p>[tex](28800 J)/(1203840 J/C )`

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