Question

Лід що мав температуру 0°С після того як його внесли до теплої кімнати перетворився на воду температура якої 20°С.У скільки разів кількість теплоти що витрачає на плавлення льоду більша від кількості теплоти яку поглинула вода під час подальшого нагрівання?

Answer 1

To answer your question, let's break it down step by step.

Given Information:

- The ice had a temperature of 0°C. - After being brought into a warm room, it transformed into water with a temperature of 20°C.

Calculating the Heat Required for Melting Ice:

To calculate the heat required for melting ice, we can use the formula:

Q = m * L

Where: - Q is the heat required (in joules) - m is the mass of the ice (in kilograms) - L is the latent heat of fusion for ice (in joules per kilogram)

The latent heat of fusion for ice is approximately 334,000 joules per kilogram.

Calculating the Heat Absorbed by the Water:

To calculate the heat absorbed by the water during further heating, we can use the formula:

Q = m * c * ΔT

Where: - Q is the heat absorbed (in joules) - m is the mass of the water (in kilograms) - c is the specific heat capacity of water (approximately 4,186 joules per kilogram per degree Celsius) - ΔT is the change in temperature (in degrees Celsius)

In this case, the change in temperature is 20°C - 0°C = 20°C.

Comparing the Amount of Heat:

To compare the amount of heat required for melting ice and the amount of heat absorbed by the water during further heating, we need to calculate the ratio:

Ratio = (Q for melting ice) / (Q for heating water)

Let's calculate the values:

1. Calculating the heat required for melting ice: - The mass of the ice is the same as the mass of the resulting water. - Let's assume the mass of the ice/water is 1 kilogram. - Using the formula Q = m * L, we get: Q for melting ice = 1 kg * 334,000 J/kg = 334,000 joules.

2. Calculating the heat absorbed by the water during further heating: - Using the formula Q = m * c * ΔT, we get: Q for heating water = 1 kg * 4,186 J/(kg°C) * 20°C = 83,720 joules.

3. Calculating the ratio: - Ratio = (Q for melting ice) / (Q for heating water) = 334,000 joules / 83,720 joules = 3.99.

Answer:

The ratio of the amount of heat required for melting the ice to the amount of heat absorbed by the water during further heating is approximately 3.99. This means that the amount of heat required for melting the ice is almost 4 times greater than the amount of heat absorbed by the water during further heating.

Please let me know if there's anything else I can help you with!