Question

Помогите!!!!! Кусок льда массой 2 кг имеет - 20 градусов Цельсия. Какое количество теплоты надо передать, чтобы превратить в воду при 20 градусов Цельсия.

Answer 1

Сперва доводим t до нуля q1=2x20x2100=21000Дж плавим q2=2x330000=660000Дж нагреваем уже воду q3=2x4200x20=168000Дж Q= q1+q2+q3=849 КДж 

Answer 2

Calculation of Heat Required to Convert Ice to Water

To calculate the amount of heat required to convert a 2 kg block of ice at -20 degrees Celsius to water at 20 degrees Celsius, we need to consider the heat required for two processes: raising the temperature of the ice from -20 degrees Celsius to 0 degrees Celsius, and then melting the ice at 0 degrees Celsius to water at 0 degrees Celsius.

1. Heat required to raise the temperature of the ice from -20 degrees Celsius to 0 degrees Celsius:

The specific heat capacity of ice is approximately 2.09 J/g°C Since we have a 2 kg block of ice, we can calculate the heat required using the formula:

Q1 = m * c * ΔT

Where: - Q1 is the heat required - m is the mass of the ice (2 kg) - c is the specific heat capacity of ice (2.09 J/g°C) - ΔT is the change in temperature (0°C - (-20°C) = 20°C)

Substituting the values into the formula, we get:

Q1 = 2 kg * 2.09 J/g°C * 20°C = 83.6 kJ

Therefore, the heat required to raise the temperature of the ice from -20 degrees Celsius to 0 degrees Celsius is 83.6 kJ.

2. Heat required to melt the ice at 0 degrees Celsius to water at 0 degrees Celsius:

The heat of fusion for ice is approximately 334 J/g Again, considering the mass of the ice (2 kg), we can calculate the heat required using the formula:

Q2 = m * ΔHf

Where: - Q2 is the heat required - m is the mass of the ice (2 kg) - ΔHf is the heat of fusion for ice (334 J/g)

Substituting the values into the formula, we get:

Q2 = 2 kg * 334 J/g = 668 J

Therefore, the heat required to melt the ice at 0 degrees Celsius to water at 0 degrees Celsius is 668 J.

Adding the heat required for both processes, we get the total heat required:

Total heat required = Q1 + Q2 = 83.6 kJ + 668 J = 83.6 kJ + 0.668 kJ = 84.268 kJ

Therefore, the total amount of heat that needs to be transferred to convert the 2 kg block of ice at -20 degrees Celsius to water at 20 degrees Celsius is approximately 84.268 kJ.

Please note that the calculations provided are based on the information available and may not account for all factors.