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 tem

Question

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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Cherry 3 years 2021-08-22T16:38:40+00:00 1 Answers 16 views 0

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  1. MichaelMet
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    2021-08-22T16:40:35+00:00
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    Answer:

    \Delta T=0.160\°C

    Explanation:

    Hello there!

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

    Q=mC\Delta T

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

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

    Best regards!

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