Question

У ванну налили 80 л води при температурі 10ºС. Скільки кг води при температурі 100ºС треба долити у ванну, щоб суміш мала температуру 25ºС?

Answer 1

Согласноф-лы подберите колл.гор.воды в кг:,чтобі совпало колл.отданого и принятого тепла
Q=cm(t1-t0) =4200*80(25-10)=5040000Дж это тепло, кот. примет хол.вода
Q=cm(t2-t1)= 4200*Х  (100-25)= -5040000 Дж  это тепло, кот.отдаст гор.вода
подберите Х и найдёте колл. гор. воды.

Answer 2

Calculation of Water Quantity and Temperature

To calculate the amount of water and the temperature required to achieve a final temperature of 25ºC in the bathtub, we can use the principle of heat transfer. The heat gained or lost by an object can be calculated using the formula:

Q = mcΔT

Where: - Q is the heat gained or lost - m is the mass of the object - c is the specific heat capacity of the substance - ΔT is the change in temperature

In this case, we have two different temperatures of water that need to be mixed. Let's assume that the initial temperature of the water in the bathtub is 10ºC and the final temperature we want to achieve is 25ºC. We also know that the initial amount of water in the bathtub is 80 liters.

To find the mass of water required at 100ºC, we can use the formula:

Q1 = mcΔT1

Where: - Q1 is the heat gained or lost by the water at 100ºC - m is the mass of the water at 100ºC - c is the specific heat capacity of water (which is approximately 4.18 J/gºC) - ΔT1 is the change in temperature from 100ºC to the final temperature of 25ºC

We can rearrange the formula to solve for the mass of water at 100ºC:

m = Q1 / (cΔT1)

To find the heat gained or lost by the water at 100ºC, we can use the formula:

Q1 = Q - Q2

Where: - Q is the total heat gained or lost by the water in the bathtub - Q2 is the heat gained or lost by the water at 10ºC

We can rearrange the formula to solve for Q1:

Q1 = Q - Q2

Now let's substitute the values into the formulas and calculate the mass of water required at 100ºC.

Calculation Steps:

1. Calculate the heat gained or lost by the water at 10ºC: - Q2 = mcΔT2 - m = 80 kg (since 1 liter of water is approximately equal to 1 kg) - c = 4.18 J/gºC - ΔT2 = 25ºC - 10ºC = 15ºC - Q2 = (80 kg) * (4.18 J/gºC) * (15ºC)

2. Calculate the total heat gained or lost by the water in the bathtub: - Q = mcΔT - m = 80 kg (since 1 liter of water is approximately equal to 1 kg) - c = 4.18 J/gºC - ΔT = 25ºC - 10ºC = 15ºC - Q = (80 kg) * (4.18 J/gºC) * (15ºC)

3. Calculate the heat gained or lost by the water at 100ºC: - Q1 = Q - Q2

4. Calculate the mass of water required at 100ºC: - m = Q1 / (cΔT1) - c = 4.18 J/gºC - ΔT1 = 100ºC - 25ºC = 75ºC

5. Convert the mass of water from grams to kilograms.

Let's calculate the values step by step:

1. Calculate the heat gained or lost by the water at 10ºC: - Q2 = (80 kg) * (4.18 J/gºC) * (15ºC)

2. Calculate the total heat gained or lost by the water in the bathtub: - Q = (80 kg) * (4.18 J/gºC) * (15ºC)

3. Calculate the heat gained or lost by the water at 100ºC: - Q1 = Q - Q2

4. Calculate the mass of water required at 100ºC: - m = Q1 / (cΔT1)

5. Convert the mass of water from grams to kilograms.

Please wait a moment while I perform the calculations.