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If 150 kcal of heat raises the temperature of 2.0 kg of a material by 400 F°, what is the specific heat capacity of the material?

0.75 kcal/kg ∙ C°
1.35 kcal/kg ∙ C°
0.19 kcal/kg ∙ C°
0.34 kcal/kg ∙ C°

Respuesta :

it would be 0.338 kcal, which would round off to 0.34.

Answer:

this result can approximate 0.34Kcal/kg.ºC

Explanation:

As during the process the material only changes its temperature we use the sensible heat formula

sensible heat formula

[tex]Q=m.ce.\Delta t[/tex]

Data

[tex]Q= 150Kcal\\m= 2.0kg\\ce=?\\\Delta t=400\ºF[/tex]

The options given to us by the unit of temperature is celsius, so we must transform from F to celsius

[tex]\ºC=(\ºF-32)/1.8[/tex]

[tex]\ºC=(400-32)/1.8\\\ºC= 204.44\ºC[/tex]

Now we clear the specific heat of the sensible heat formula

[tex]Q=m.ce.\Delta t\\\frac{Q}{m\Delta t}=ce\\\frac{150Kcal}{2Kg.204.44\ºC}=ce\\0.36Kcal/g.\ºC[/tex]

this result can approximate 0.34Kcal/kg.ºC

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