Question

Лед массой 2 кг, при температуре -10 градусов вынесли в комнату , после чего лед растаял а образовавшаяся вода нагрелась до 18 градусов . Сколько теплоты для этого потребовалась?

Answer 1

С=удельная теплоемкость льда 2100 Дж/кг*С
t1-начальная темп. льда. =-10градусов
t2-конечная темп.  18 градусов.
 m=2кг.
Q=c*m*(t2-t1)=2100*2*(18-(-10))=117.600ДЖ

Answer 2

Calculation of Heat Required to Melt Ice and Heat Water

To calculate the amount of heat required to melt the ice and heat the resulting water, we need to consider the specific heat capacity of ice and water, as well as the latent heat of fusion for ice.

1. Specific Heat Capacity of Ice: The specific heat capacity of ice is the amount of heat required to raise the temperature of 1 kilogram of ice by 1 degree Celsius. According to the available search results, the specific heat capacity of ice is not provided directly. However, we can assume it to be approximately 2.09 J/g°C based on general knowledge.

2. Latent Heat of Fusion for Ice: The latent heat of fusion for ice is the amount of heat required to convert 1 kilogram of ice at its melting point to water at the same temperature. The latent heat of fusion for ice is approximately 334 J/g.

3. Temperature Change: The initial temperature of the ice is -10 degrees Celsius, and it is heated to 18 degrees Celsius. Therefore, the temperature change is 18 - (-10) = 28 degrees Celsius.

4. Mass of Ice: The mass of the ice is given as 2 kilograms.

Calculation:

To calculate the heat required to melt the ice and heat the resulting water, we need to consider two steps:

1. Heat required to melt the ice: The heat required to melt the ice can be calculated using the formula: Heat = mass of ice * latent heat of fusion for ice

Substituting the values: Heat = 2 kg * 334 J/g = 668 J

2. Heat required to heat the water: The heat required to heat the water can be calculated using the formula: Heat = mass of water * specific heat capacity of water * temperature change

Since the ice has melted and turned into water, the mass of water is equal to the mass of ice, which is 2 kilograms.

Substituting the values: Heat = 2 kg * 4.18 J/g°C * 28°C = 234.24 J/g * 2 kg * 28°C = 13107.84 J

Total Heat Required:

To find the total heat required, we add the heat required to melt the ice and the heat required to heat the water:

Total Heat Required = Heat required to melt the ice + Heat required to heat the water Total Heat Required = 668 J + 13107.84 J = 13775.84 J

Therefore, 13775.84 Joules of heat are required to melt the ice and heat the resulting water.

Please note that the specific heat capacity of ice and the latent heat of fusion for ice can vary slightly depending on the source and the purity of the ice. The values used in this calculation are approximate averages.