Question

1. Exactly 1000 J of energy as heat are added to an 8.50 g sample of helium at 20.0 °C under constant-pressure. The specific heat capacity at constant-pressure for helium is 5.1926 kJ kg^-1 K^-1. What is the final temperature of the helium sample?
   1. 316 K
   2. 22.7 K
   3. 273 K
   4. 89.6 K
   5. 363 K

Answer 1

Amount of heat absorbed , Q = mcdT

Where

m = mass of Helium = 8.50g

c = specific heat capacity at constant-pressure for helium = 5.1926 kJ kg^-1 K^-1.

dT = change in temperature = final - initial

    = T - (20.0oC)

    = T - ( 20.0+273)K

Q = amount of heat added = 1000 J

Substitute the above values we get dT = Q / (mc)

                                                       = 1000J / (8.50g * 5.1926 kJ kg^-1 K^-1)

                                                       = 22.6 K

                             T - ( 20.0+273)K = 22.6 K

                                                   T = 22.6K+(20.0+273)K

                                                      = 315.6

                                                      ~ 316 K

Therefore option (a) is correct