Question

Какое количество теплоты необходимо отнять у 200 г воды, нагретой до 30 градусов по Цельсию, чтобы получить лед при температуре -5 градусов по Цельсию вопрос

Answer 1

4200*0,2*30+330000*0,2+2100*0,2*5=25200+66000+2100=93300Дж=93,3кДж

Answer 2

Calculating Heat Transfer for Water Turning into Ice

To calculate the amount of heat needed to be removed from 200g of water at 30°C to turn it into ice at -5°C, we can use the formula:

Q = mcΔT + mL

Where: - Q = heat energy - m = mass of the substance - c = specific heat capacity - ΔT = change in temperature - L = latent heat of fusion

Given: - m = 200g - t1 = 30°C - t2 = -5°C - c = specific heat capacity of water - L = latent heat of fusion of water

Calculation Steps

1. Calculate the heat required to lower the temperature of water from 30°C to 0°C. 2. Calculate the heat required to freeze the water at 0°C.

Heat Transfer Calculation

1. Heat to lower the temperature to 0°C: - Using the formula Q = mcΔT, where c = specific heat capacity of water (4.18 J/g°C) and ΔT = t2 - t1: - Q1 = mcΔT - Q1 = (200g)(4.18 J/g°C)(0°C - 30°C) - Q1 = -25080 J

- Citation: Source 2. Heat to freeze the water at 0°C: - Using the formula Q = mL, where L = latent heat of fusion of water (334 J/g): - Q2 = mL - Q2 = (200g)(334 J/g) - Q2 = 66800 J

- Citation: Source

Total Heat Transfer

The total heat transfer required is the sum of Q1 and Q2: - Total heat transfer (Q) = Q1 + Q2 - Total heat transfer (Q) = -25080 J + 66800 J - Total heat transfer (Q) = 41720 J

Therefore, the amount of heat that needs to be removed from 200g of water, heated to 30°C, to obtain ice at -5°C is 41720 Joules.

Final Answer

The amount of heat that needs to be removed from 200g of water, heated to 30°C, to obtain ice at -5°C is 41720 Joules.