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?

Answer 1

`\Delta T=0.160\°C`

Step-by-step 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=(Q)/(mC)\\\\ \Delta T=(125.7cal)/(634.5g*1.241 cal/(g\°C)) \\\\ \Delta T=0.160\°C`

Best regards!