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

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

Answer 2

this result can approximate 0.34Kcal/kg.ºC

Step-by-step 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\\(Q)/(m\Delta t)=ce\\(150Kcal)/(2Kg.204.44\ºC)=ce\\0.36Kcal/g.\ºC`

this result can approximate 0.34Kcal/kg.ºC