# Calculating Heat Required to Change 1.0 kg of Ice to Steam

• Mr. Goosemahn
In summary, the problem can be divided into five sections: finding the heat required to raise the temperature of the ice (Q1), the heat required to melt the ice (Q2), the heat required to raise the temperature of the water (Q3), the heat required to vaporize the water (Q4), and the heat required to raise the temperature of the steam (Q5). Using the given values for the constants, these equations can be solved to find the total heat required to change 1.0 kg of ice at -30 degrees C into steam at 120 degrees C under 1.0 atm of pressure.
Mr. Goosemahn

## Homework Statement

How much heat in kcal is required to change 1.0 kg of ice, originally at -30 degrees C, into steam at 120 degrees C? Assume 1.0 atm of pressure. (Please use the following values for the constants needed in the problem: Latent heat of fusion=79.7 kcal/kg; Specific heat of ice=0.5 kcal/kg/K; Latent heat of vaporization=539 kcal/kg; Specific heat of water 1.0 kcal/kg/K; Specific heat of steam=0.480 kcal/kg)

## Homework Equations

Qmelt=mLfusion
Qfreeze=-mLfusion
Qvaporization=mLvaporization
Qcondensation=-mLvaporization

## The Attempt at a Solution

Quite honestly, I don't even know how to start. I don't see how the 1.0 atm of pressure is involved in the equations.

Could somebody just tell me how to start this off and I'll take it from there.

I've been reading from my book but it's not clear.

I had originally tried to use the equation Q=cmT, with the values plugged in as Q=4186*1*150, giving me 627.9 kilocalories, but this is incorrect.

Divide the problen into five sections.
i) Q1 = m*ci*ΔT----(1)
ii) Q2 = m*Lfreeze -----(2)
similarly right the remaining equations and find the total Q..

This is because the equation Q=cmT only applies to a change in temperature, not a change in state. In order to calculate the heat required for a change in state, we need to use the specific latent heat equations.

To start, we need to find the heat required to melt the ice at -30 degrees C. This can be calculated using the equation Qmelt=mLfusion, where m is the mass of the ice (1.0 kg) and Lfusion is the latent heat of fusion (79.7 kcal/kg). Plugging in these values, we get Qmelt=1.0*79.7=79.7 kcal.

Next, we need to find the heat required to raise the temperature of the water from -30 degrees C to 100 degrees C. This can be calculated using the equation Q=cmT, where c is the specific heat of ice (0.5 kcal/kg/K) and T is the change in temperature (100-(-30)=130 K). Plugging in these values, we get Q=0.5*1.0*130=65 kcal.

After the ice has melted, we need to find the heat required to vaporize the water at 100 degrees C. This can be calculated using the equation Qvaporization=mLvaporization, where m is the mass of the water (1.0 kg) and Lvaporization is the latent heat of vaporization (539 kcal/kg). Plugging in these values, we get Qvaporization=1.0*539=539 kcal.

Finally, we need to find the heat required to raise the temperature of the steam from 100 degrees C to 120 degrees C. This can be calculated using the equation Q=cmT, where c is the specific heat of steam (0.480 kcal/kg) and T is the change in temperature (120-100=20 K). Plugging in these values, we get Q=0.480*1.0*20=9.6 kcal.

To find the total heat required, we need to add up all of these values: 79.7+65+539+9.6=693.3 kcal.

Therefore, the heat required to change 1.0 kg of ice at -30 degrees C to steam at 120 degrees C is 693.3 kcal. The 1.0 atm of pressure is not directly involved in the calculations,

## 1. How do you calculate the heat required to change 1.0 kg of ice to steam?

The heat required to change 1.0 kg of ice to steam can be calculated using the formula Q = m x L, where Q is the heat required, m is the mass of the substance, and L is the latent heat of transformation. For water, the latent heat of fusion (from ice to water) is 334 kJ/kg and the latent heat of vaporization (from water to steam) is 2260 kJ/kg.

## 2. What is the specific heat of ice and steam?

The specific heat of ice is 2.108 J/g°C and the specific heat of steam is 1.996 J/g°C. This means that it takes more energy to raise the temperature of ice compared to steam.

## 3. How much heat is required to change 1.0 kg of ice at 0°C to steam at 100°C?

The total heat required is the sum of the heat to melt the ice and the heat to vaporize the water. Using the formula Q = m x L, the heat required for melting 1.0 kg of ice is 334 kJ and the heat required for vaporizing 1.0 kg of water is 2260 kJ. Therefore, the total heat required is 2594 kJ.

## 4. Is there a difference in the heat required to change 1.0 kg of ice to steam at different temperatures?

Yes, the heat required to change 1.0 kg of ice to steam at different temperatures will vary. This is because the specific heat and latent heat of transformation for water changes with temperature. As the temperature increases, the specific heat decreases and the latent heat of transformation increases.

## 5. Can you use the same formula to calculate the heat required for other substances?

No, the formula Q = m x L can only be used to calculate the heat required for substances that change from one phase to another (e.g. ice to steam). For substances that change temperature within the same phase, the formula Q = m x C x ΔT is used, where C is the specific heat and ΔT is the change in temperature.

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