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A 634.5 g sample of helium absorbs 125.7 calories of heat. The specific heat capacity of helium is 1.241 cal/(g·°C). By how much did the temperature of this sample change, in degrees Celsius?

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Sebassandinidn

Answer:

[tex]\Delta T=0.160\°C[/tex]

Explanation:

Hello there!

In this case, according to the following equation for the calculation of heat in this calorimetry problem:

[tex]Q=mC\Delta T[/tex]

It is possible for us to calculate to calculate the change in temperature for this process by solving for DT in the aforementioned equation:

[tex]\Delta T=\frac{Q}{mC}\\\\ \Delta T=\frac{125.7cal}{634.5g*1.241 cal/(g\°C)} \\\\ \Delta T=0.160\°C[/tex]

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