Heat Absorbed Changing 2kg Ice to Steam: Calculation

[panther7]

How much heat is absorbed in changing 2-kg of ice at -5°C to steam at 110.0°C? The Specific heats of ice, liquid water, and steam are, respectively, 2060 J/Kg x K, 4180 J/Kg x K, and 2020 J/Kg x K. The heat fusion of ice is 3.34 x 10^5 J/Kg. The heat of vaporization of water is 2.26 x 10^6 J/kg

 

[russ_watters]

Is this homework? You have all the info you need there - what have you tried?

 

[astrokreng]

To calculate the heat absorbed in changing 2kg of ice at -5°C to steam at 110.0°C, we can use the formula Q = m * c * ΔT, where Q is the heat absorbed, m is the mass, c is the specific heat, and ΔT is the change in temperature.

First, we need to calculate the heat absorbed in changing the ice to water. The ice is at -5°C, so ΔT = 5°C. The specific heat of ice is 2060 J/Kg x K, so we can calculate the heat absorbed as follows:

Q1 = m * c * ΔT = 2kg * 2060 J/Kg x K * 5°C = 20600 J

Next, we need to calculate the heat absorbed in changing the water to steam. The water is at 0°C and we want to reach 110.0°C, so ΔT = 110.0°C. The specific heat of liquid water is 4180 J/Kg x K, so we can calculate the heat absorbed as follows:

Q2 = m * c * ΔT = 2kg * 4180 J/Kg x K * 110.0°C = 918800 J

Lastly, we need to account for the heat of fusion and heat of vaporization. The heat of fusion of ice is 3.34 x 10^5 J/Kg and the heat of vaporization of water is 2.26 x 10^6 J/Kg. Since we are changing 2kg of ice to steam, we need to add these two values together:

Q3 = 2kg * (3.34 x 10^5 J/Kg + 2.26 x 10^6 J/Kg) = 5.92 x 10^6 J

Therefore, the total heat absorbed in changing 2kg of ice at -5°C to steam at 110.0°C is:

Q = Q1 + Q2 + Q3 = 20600 J + 918800 J + 5.92 x 10^6 J = 6.93 x 10^6 J

In conclusion, the heat absorbed in changing 2kg of ice at -5°C to steam at 110.0°C is 6.93 x 10^6 J. This calculation takes into account the specific heats of