Temperature/Heat Problem?

Question

Hot water at 90°C is poured into a large Styrofoam cup that contains water at 20°C. The final temperature of the mixture is 75°C, closer to the initial temperature of the hot water.

a.) Which mass of water was the greatest, the hot or the cold water? Does that mean that the heat lost by the water is less than the heat gained by cold water? Explain.

b.) Calculate the ratio of cold water to hot that would give this result. Start with energy conservation: Qн + Qс = 0.

c.) If there was 150g of hot water calculate how much heat was exchanged?

Answer 1

Ok, let's go step by step.

You know, by theory that the heat the cold water has gained, is the same as the heat lost by the hot water.

In other words, the heat lost by the water is equal in numbers than the heat gained by cold water.

To calculate the ratio of masses :

ma = mass of hot water

mb = mass of cold water.

Q = ma*1*(90-75)

Q = 15ma

Q' = mb*1*(75 - 20)

Q' = 55mb

15 ma = 55mb

ma / mb = 11 / 3

mass of hot water is greater than mass of cold water.

b) The ratio is the one that I have already found :

ma / mb = 11 / 3

c) ma = 150 g, then to fin the heat exchanged, let's just find the heat lost by the hot water :

150*15 = 2250 cal

Hope that helps

Answer 2

Qt=Q1+Q2
Q1=Cp m1 (T2-T1)
Q2=Cp m2 (T3-T2)

T2=75C T1=20C
T3=90C

(a) mass of the water at 90C was greater since the temperature gradient is less between 90 C then it is 20C

(b) Q1/Q3= m1 (T2-T1)/ m2 (T3-T2)
Q1/Q3=(m1/m2)(75-20)/(90-75)
Q1/Q3=(55/15)(m1/m2)
and Qн + Qс not equal 0
Qн + Qс = Q (final) unless I missed something

c)Cp = 4.1813Joules/( g K)
Q2=Cp 150 (90-75)=4.1813 x 150 x 15=9408 joules

Answer 3

probably no longer.. like bypass out as in feint? or like get all stoned and bypass out in mattress? its probably oftentimes used, i exceeded out from getting a haircut and it in actuality exceeded off as immediately as, through no potential decrease back.