Question

У прозорий калориметр, що містить 650см3 води, занурюють електричну лампочку накалювання потужністю 60 Вт. За три хвилини вода нагрівасться на 3,5 °С. Яка частина енергії, що витрачасться, пропускається калориметром назовні у вигляді променистої енергіт? ​

Answer 1

Ответ:

650см3 води, занурюють електричну лампочку накалювання потужністю 60 Вт.

Answer 2

Calculation of Energy Transferred as Radiant Energy

To determine the fraction of energy that is transferred as radiant energy outside the calorimeter, we need to calculate the total energy transferred to the water and then subtract the energy absorbed by the water.

The formula to calculate the energy transferred to the water is:

Q = mcΔT

Where: - Q is the energy transferred to the water (in joules) - m is the mass of the water (in kilograms) - c is the specific heat capacity of water (approximately 4.18 J/g°C or 4.18 J/cm³°C) - ΔT is the change in temperature of the water (in °C)

Given: - Volume of water (V) = 650 cm³ - Mass of water (m) = V × density of water (approximately 1 g/cm³) - Power of the lamp (P) = 60 W - Time (t) = 3 minutes = 180 seconds - Change in temperature (ΔT) = 3.5 °C

Let's calculate the energy transferred to the water:

m = V × density of water = 650 cm³ × 1 g/cm³ = 650 g = 0.65 kg

Q = mcΔT = 0.65 kg × 4.18 J/g°C × 3.5 °C = 9.0455 J

Therefore, the energy transferred to the water is approximately 9.0455 joules.

Calculation of Energy Transferred as Radiant Energy

To calculate the energy transferred as radiant energy outside the calorimeter, we need to subtract the energy absorbed by the water from the total energy transferred to the water.

Given that the power of the lamp is 60 W and the time is 3 minutes (180 seconds), we can calculate the total energy transferred to the water:

Total energy transferred = Power × Time = 60 W × 180 s = 10800 J

Now, let's calculate the energy absorbed by the water:

Energy absorbed by the water = Energy transferred to the water = 9.0455 J

Finally, we can calculate the energy transferred as radiant energy outside the calorimeter:

Energy transferred as radiant energy = Total energy transferred - Energy absorbed by the water = 10800 J - 9.0455 J = 10790.9545 J

Therefore, approximately 10790.9545 joules of energy are transferred as radiant energy outside the calorimeter.

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