Question

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°

Answer 1

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

$Q=m.ce.\Delta t$

Data

$Q= 150Kcal\\m= 2.0kg\\ce=?\\\Delta t=400\ºF$

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

$\ºC=(\ºF-32)/1.8$

$\ºC=(400-32)/1.8\\\ºC= 204.44\ºC$

Now we clear the specific heat of the sensible heat formula

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

this result can approximate 0.34Kcal/kg.ºC

Answer 2

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