Find the heat necessary to change 2.5kg of water (ice) at -2 degrees C to steam at 105 degrees C.

specific heat ice= 2060
specific heat water= 4180
specific heat steam= 2020
heat of fusion= 3.34X10^5
heat of vaporization= 2.26X10^6

One mole of diatomic ideal gas is carried around the cycle shown in fig.2. The temperature and pressure at state D are 300-k and 1.01*10

To find the heat necessary to change 2.5kg of water (ice) at -2 degrees C to steam at 105 degrees C, we need to consider the different phases of water and calculate the heat required for each phase change.

First, let's determine the heat required to raise the temperature of ice from -2 degrees C to 0 degrees C and then melt it into water at 0 degrees C.

The heat required to raise the temperature of ice can be calculated using the formula:

Q1 = mass × specific heat ice × change in temperature

Q1 = 2.5kg × 2060 J/(kg·°C) × (0 - (-2)°C)

Q1 = 2.5kg × 2060 J/(kg·°C) × 2°C

Q1 = 10300 J

Next, we need to calculate the heat required to melt the ice into water at 0 degrees C using the formula:

Q2 = mass × heat of fusion

Q2 = 2.5kg × 3.34 × 10^5 J/kg

Q2 = 835000 J

Now, let's calculate the heat required to raise the temperature of the water from 0 degrees C to 100 degrees C:

Q3 = mass × specific heat water × change in temperature

Q3 = 2.5kg × 4180 J/(kg·°C) × (100 - 0)°C

Q3 = 2.5kg × 4180 J/(kg·°C) × 100°C

Q3 = 1.045 × 10^6 J

Finally, let's calculate the heat required to change water at 100 degrees C into steam at 105 degrees C:

Q4 = mass × heat of vaporization

Q4 = 2.5kg × 2.26 × 10^6 J/kg

Q4 = 5.65 × 10^6 J

Now, we can sum up all the calculated heats to find the total heat necessary:

Total heat = Q1 + Q2 + Q3 + Q4

Total heat = 10300 J + 835000 J + 1.045 × 10^6 J + 5.65 × 10^6 J

Total heat = 7.533 × 10^6 J

Therefore, the heat necessary to change 2.5kg of water (ice) at -2 degrees C to steam at 105 degrees C is 7.533 × 10^6 Joules.