Question

a vessel of copper of mass 0.21 kg contains 20 g of water at 80 degree celcius if 100 g water at 10 degree celsius is poured into it find the final temperature of mixture​

Answer 1

Answer:

Constants that we need:

Heat capacity of water is Cpw = 4.186J/g-K (same for vessel)

The final temperature of the water is denoted as Tf

To raise the temperature of this 80g + 20g = 100g of water and vessel from 20C to Tf requires

ΔH1 = mCpwΔT = 100(4.186)(Tf – 20) = (418.6Tf – 8372)J

To cool 100g of water from 40C to Tf releases

ΔH2 = mCpwΔT = 100(4.186)(40 – Tf) = (16,744 – 418.6Tf)J

But ΔH1 = ΔH2 in a sealed system (no heat flow to or from the outside)

418.6Tf – 8372 = 16744 – 418.6Tf

837.2Tf = 25,116

Tf = 30

The final temperature of the water and vessel is 30C.

P.S. We could just have said that 100g at 20C + 100g at 40C results in 200g at 30C, but the method shown is the correct way to calculate the temperature.

Explanation:

MARK AS BRAINLIEST

Answer 2

Concept:

• Heat capacity
• A physical feature of matter known as heat capacity or thermal capacity is the quantity of heat that must be applied to an object in order to cause a unit change in temperature.

Given:

• Mass of copper m= 0.21 kg = 210 g
• Mass water in the vessel = 20 g
• Initial temperature T1 = 80 °C
• Mass of water added = 100 g
• The temperature of the water added T2 = 10°C

Find:

• The final temperature of the mixture

Solution:

Q = mcΔT

Let the final temperature of the mixture = Tf

The heat exchanged is the same

Heat capacity of water c = 4.186J/g-K

Heat capacity of copper c = 0.385J/g-K

Q1 = 20 (4.186) (80-Tf) + 210 (0.385) (80-Tf)

Q1 = 83.72 (80-Tf) + 80.85 (80-Tf)

Q1 = 164.57 (80-Tf)

Q2 = 100 (4.186) (Tf-10)

Q2 = 418.6 (Tf-10)

Q1 = Q2

164.57 (80-Tf) = 418.6 (Tf-10)

13165.6 - 164.57 Tf = 418.6 Tf - 4186

17351.6 = 583.17 Tf

Tf = 17351.6/583.17

Tf = 29.7 °C

The final temperature of the mixture is 29.7 °C.

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