Question

Какое количество теплоты получила вода при нагревании от 15 до 25 градусов в бассейне длина которого 100 метров ширина 6 метров и глубина 2 метра?

Answer 1

Q=cpV(t2-t1)
Q=4200*1000*100*6*2*10=5.04*10^10Дж

Answer 2

Calculation of Heat Energy Absorbed by Water

To calculate the amount of heat energy absorbed by water when it is heated from 15 to 25 degrees Celsius in a pool with dimensions of 100 meters in length, 6 meters in width, and 2 meters in depth, we can use the formula:

Q = mcΔT

Where: - Q is the heat energy absorbed by the water (in joules) - m is the mass of the water (in kilograms) - c is the specific heat capacity of water (in joules per kilogram per degree Celsius) - ΔT is the change in temperature (in degrees Celsius)

To find the mass of the water, we can use the formula:

m = Vρ

Where: - V is the volume of the water (in cubic meters) - ρ is the density of water (in kilograms per cubic meter)

Let's calculate the heat energy absorbed by the water step by step.

Step 1: Calculate the Volume of Water

The volume of the pool can be calculated by multiplying its length, width, and depth:

Volume = Length × Width × Depth

Volume = 100 m × 6 m × 2 m = 1200 cubic meters

Therefore, the volume of water in the pool is 1200 cubic meters.

Step 2: Calculate the Mass of Water

The density of water is approximately 1000 kilograms per cubic meter.

Mass = Volume × Density

Mass = 1200 m^3 × 1000 kg/m^3 = 1,200,000 kilograms

Therefore, the mass of water in the pool is 1,200,000 kilograms.

Step 3: Calculate the Change in Temperature

The change in temperature is given as 25 degrees Celsius minus 15 degrees Celsius:

ΔT = 25 °C - 15 °C = 10 degrees Celsius

Step 4: Calculate the Specific Heat Capacity of Water

The specific heat capacity of water is approximately 4.18 joules per gram per degree Celsius, or 4180 joules per kilogram per degree Celsius.

Therefore, the specific heat capacity of water is 4180 joules per kilogram per degree Celsius.

Step 5: Calculate the Heat Energy Absorbed by the Water

Now we can use the formula Q = mcΔT to calculate the heat energy absorbed by the water:

Q = (mass of water) × (specific heat capacity of water) × (change in temperature)

Q = 1,200,000 kg × 4180 J/(kg·°C) × 10 °C

Calculating this value gives us:

Q = 50,160,000,000 joules

Therefore, the water in the pool absorbed approximately 50,160,000,000 joules of heat energy when it was heated from 15 to 25 degrees Celsius.

Please note that the above calculation assumes that the specific heat capacity of water remains constant over the given temperature range. In reality, the specific heat capacity of water can vary slightly with temperature, but for this calculation, we have used the average value.