5 g ce of num (which has a mol heat capacity of0.89 Jl C g healed to 82.4 Cand d

Question

5 g ce of num (which has a mol heat capacity of0.89 Jl C g healed to 82.4 Cand dropped into a calo eler contai g wal ater in the calorimet A 212 g D 5.72 kg C 5.42 g o D 168 kg none of these QUESTION 20 0.0 mL of pure 282 K m d 50.0 mL ure water at 306 K. What is the 1nal emperature the mixtul A 342K B. 588 K C 297 K D.294 K E.24 QUESTION 2 The Slu Is the B.newol C kilojo E-ampere QUESTION 22 P 1 am and T 370 K? Whi the followi tatements correctly d for he following othermic process and H209) H206) o A q and w are both zero. negative, Wi Ca and E.qis p QUESTION 23 of 100 liter to at 8.93 liters ot 1.00 am How A 402 B.003 o C241 103 0.905 J E.ncne c Utese heat capacity of wa ch work 4.18 JMWg C) 23 C. mperature ricantig the wat 24.2 ignoring significa calculate the 1 points Save Answer 1 points Save Answwr T 300 K

Explanation / Answer

Ans.20

LEt the final temperature be x.

Heat lost = Heat gained

Heat = mcdeltaT = m(mass),c(heat capacity)deltaT(final temp-initial temp)

Density of.pure water = 1g/cm3

Mass = Volume*density

Mass=Volume

30g*c*(x-282)K = 50g*c*(306-x)K

30/50 = 306-x/x-282

3x - 846 = 1530-5x

8x = 2376

x = 297K

Final temperature of the mixture = 297K

21.SI unit of pressure is Pascal.

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