Question

Two hundred grams of a substance requires 18.8 kj of heat to raise its temperature from 25°C to 45°C. Use the table to identify the substance.

Answer 1

11) q = (m)(Cp)(final temperature- initial temperature) 520 J = (200 g)(cp)(20)---->since the heat capacity is in JOULES convert KJ to Joules so .52 kj = 520 J .13 = Cp, which is gold, so D is your answer.

12) q = (20)(334) q = 6,680 J q = 6.68 KJ, which is D, so D is your answer.

13) q = (m)(cp)(final temperature-initial temperature) q = (150)(.92)(15) q = 2070 J q = 2.07 KJ, which is B, so B is your answer.

18) q = (3.16)(.50) q = 1.58 KJ, which is C, so C is your answer.

19) q = (.13 mol)(40.67 kj/mol) q = 565 KJ, which is D, so D is your answer. I hope this helps!

Answer 2

Answer: ammonia

Explanation:

`Q= m* c* \Delta T`

Q= heat gained = 18.8 kJ

m= mass of the substance = 200 g

c = heat capacity of unknown substance= ? J/g ° C

`\Delta T={\text{Change in temperature}}=(45-25)^0C=20^oC`

`18800J=200g* cJ/g^oC*20^oC`

`c=4.7J/g^0C`

Thus the unknown substance is ammonia which has a heat capacity of
`4.7J/g^0C`