Question

one little of water at 100c is added to four litres of water at 30c.what will be the final temperature of the water?

Answer 1

The final temperature of the water will be approximately 46.25°C.

Step-by-step explanation:

Let's call the initial temperature of the first liter of water T1 = 100°C and the initial temperature of the four liters of water T2 = 30°C.

The total mass of the water is 1 + 4 = 5 liters.

Using the equation of Heat Transfer: Q = mcΔT, where Q is the heat transferred, m is the mass, c is the specific heat capacity of water, and ΔT is the change in temperature, we can calculate the heat transfer from the first liter of water to the four liters of water.

Q = (1 kg) * (4.18 J/g°C) * (100°C - T3)

Q = (4 kg) * (4.18 J/g°C) * (T3 - 30°C)

Solving for T3, we get:

T3 = 46.25°C

So, the final temperature of the water will be approximately 46.25°C.

Answer 2

o calculate the final temperature of the water, we can use the formula for heat transfer:

Q = mcΔT

where Q is the heat transfer, m is the mass of the water, c is the specific heat capacity, and ΔT is the change in temperature.

For the one liter of water at 100°C, the heat transfer is:

Q1 = (1 kg) * (4.18 J/g°C) * (100°C - 30°C) = 4180 J

For the four liters of water at 30°C, the heat transfer is:

Q2 = (4 kg) * (4.18 J/g°C) * (100°C - 30°C) = 16680 J

Since heat is conserved, the total heat transfer must remain constant:

Q1 + Q2 = Qfinal

Therefore, the final temperature of the water can be calculated as:

Qfinal = (5 kg) * (4.18 J/g°C) * ΔTfinal

Solving for ΔTfinal, we get:

ΔTfinal = Qfinal / (5 kg) * (4.18 J/g°C) = (Q1 + Q2) / (5 kg) * (4.18 J/g°C) = (4180 J + 16680 J) / (5 kg) * (4.18 J/g°C) = (20860 J) / (5 kg) * (4.18 J/g°C) = 40°C

Therefore, the final temperature of the water will be 40°C.

Step-by-step explanation: