A500-ml bottle of water at room temperature and a 2-l bottle of water at the same temperature were placed in a refrigerator. after 30 minutes, the 500-ml bottle of water had cooled to the temperature of the refrigerator. an hour later, the 2-l of water had cooled to the same temperature. when asked which sample of water lost the most heat, one student replied that both bottles lost the same amount of heat because they started at the same temperature and finished at the same temperature. a second student thought that the 2-l bottle of water lost more heat because there was more water. a third student believed that the 500-ml bottle of water lost more heat because it cooled more quickly. a fourth student thought that it was not possible to tell because we do not know the initial temperature and the final temperature of the water. indicate which of these answers is correct and describe the error in each of the other answers.

The answer to your question is Second student

Explanation:

Let's pretend the data are:

500-ml bottle                        2-L bottle

T1 = 25°C                               T1 = 25°C

T2 = 4°C                                T2 = 4°C

Cp = 4.18 J/g°C

Density = 1 g/ml

then

mass 1 = 500 g                     mass 2 = 2000 g

Gradient of temperatures

ΔT = 4 - 25 = -21°C               ΔT = 4 - 25 = -21°C

Heat formula

Q = mCpΔT

Process

Calcule heat gained or lost in both situations

Q1 = (500)(4.18)((-21)            Q2 = (2000)(4.18)(-21)

Q1 = -43890 J                      Q2 = -175560 J

Conclusion: The second bottle lost more heat.

The first student was wrong because the heat lost depend on the mass quantity not the the initial or final temperature.

The second student was right, heat depends on the quantity of matter.

The third student was wrong because heat lost do not depend on the speed it is lost or gained.

The fourth student is wrong because we can stablish temperature as constant, and then only quantity of matter changes.

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g