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A cold pack containing 200 grams of ice at 0 degrees celcius is placed on an athlete's shoulder. When the pack is removed the temperature of the liquid water is 40 degrees celcius. How much energy in KJ was absorbed by the cold pack?

Respuesta :

Answer:

366.99 kJ

Explanation:

Enthalpy of melting of ice = 333.55 kJ

The expression for the calculation of the enthalpy change of a process in which liquid water at 0 °C converts to 40 °C is shown below as:-

[tex]\Delta H=m\times C\times \Delta T[/tex]

Where,  [tex]\Delta H[/tex]  is the enthalpy change

m is the mass

C is the specific heat capacity

[tex]\Delta T[/tex]  is the temperature change

Thus, given that:-

Mass = 200 g

Specific heat = 4.18 J/g°C

[tex]\Delta T=40-0\ ^0C=40\ ^0C[/tex]

So,  

[tex]\Delta H=200\times 4.18\times 40\ J=33440\ J[/tex]

Also, 1 J = 0.001 kJ

So,  

[tex]\Delta H=33.44\ kJ[/tex]

So, total energy absorbed = 333.55 + 33.44 = 366.99 kJ

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