Thermodynamics heat absorption problem

[panther7]

Homework Statement

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

Homework Equations

Q=mHf
Q=mHv
Q=mC(change in temp)

The Attempt at a Solution

?

 

[dynamicsolo]

You have all the information and formulas you need. Consider that you have five stages that the water has to go through in this process:

the ice must warm up from -5º C to 0º C ;
the ice must then melt into (liquid) water at a constant 0º C ;
the meltwater must now warm from 0º C to 100º C ;
the water must next vaporize into steam at a constant 100º C ; and, finally,
the steam must heat up from 100º C to 110º C .

So you will have five heat terms to compute, using the given information (they were nice to you and didn't even make you do unit conversions). You will get a total heat input in Joules.

 

[Moetasim]

I would first identify the key information and variables provided in the problem. The mass of the substance (2 kg) and the temperatures (-5°C and 110.0°C) are given, as well as the specific heats and heat fusion and vaporization values for ice and water. These values will be crucial in solving the problem.

Next, I would use the equations Q=mHf and Q=mHv to calculate the heat absorbed in changing the ice to water and the water to steam, respectively. Using the given values, I would plug them into the equations to get:

Q1 = (2 kg)(3.34 x 10^5 J/kg) = 6.68 x 10^5 J
Q2 = (2 kg)(2.26 x 10^6 J/kg) = 4.52 x 10^6 J

Then, I would use the equation Q=mC(change in temp) to calculate the heat absorbed in raising the temperature of the water from 0°C to 110.0°C:

Q3 = (2 kg)(4180 J/Kg x K)(110.0°C - 0°C) = 9.18 x 10^5 J

Finally, I would add all three values to get the total heat absorbed:

Qtotal = Q1 + Q2 + Q3 = (6.68 x 10^5 J) + (4.52 x 10^6 J) + (9.18 x 10^5 J) = 6.19 x 10^6 J

Therefore, the total heat absorbed in changing 2-kg of ice at -5°C to steam at 110.0°C is 6.19 x 10^6 J.