How Much Heat is Needed to Convert Ice at -10C to Steam at 110C?

[bommie]

How much heat (in joules) must be added to change 20.0g of H₂O(s) at -10C to steam at 110C? The specific heat of H₂O(s) is 2.09 J/gC; of H₂O(L) is 4.184 J/gC; and of H₂O(g) is 1.84 J/gC. The heat fusion of water is 334.7 J/g and the heat vaporization of water is 2259.4J/g.I feel like we have to find the temp for s ---> l and then l ---> g. But I'm completely lost; I don't understand Thermochemistry.

 

[Borek]

Split it into separate parts - you have to warm up ice till it can be melted, you have to melt it, you have to warm the the water till boiling point, you have to boil it, you finally have to warm up steam. Then add everything.

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methods

 

[Blueshift5]

Sure, I can help you with this problem. Thermochemistry deals with the study of heat and energy changes in chemical reactions. In this case, we need to calculate the amount of heat required to change 20.0g of H2O(s) at -10C to steam at 110C.

First, we need to break down the steps involved in this process. We have to first change the solid ice at -10C to liquid water at 0C, and then further heat the liquid water to reach a temperature of 110C, where it will turn into steam. Each of these steps requires a specific amount of heat, which we can calculate using the given specific heat values and heat fusion/vaporization values.

For the first step, we need to calculate the amount of heat required to change the solid ice at -10C to liquid water at 0C. This can be done using the formula Q = m * ΔT * specific heat. Here, m represents the mass of the substance, ΔT represents the change in temperature, and specific heat is the specific heat of the substance. Plugging in the values, we get Q = (20.0g) * (0C - (-10C)) * (2.09 J/gC) = 418 J.

Next, we need to calculate the amount of heat required to heat the liquid water from 0C to 110C. This can be done using the same formula, but this time we will use the specific heat of liquid water, which is 4.184 J/gC. Q = (20.0g) * (110C - 0C) * (4.184 J/gC) = 9180.8 J.

Finally, we need to calculate the amount of heat required to change the liquid water at 110C to steam at 110C. This step involves the heat vaporization of water, which is the amount of heat required to convert liquid water into steam at the same temperature. We can calculate this using the formula Q = m * heat vaporization. Plugging in the values, we get Q = (20.0g) * (2259.4 J/g) = 45188 J.

Now, to find the total amount of heat required, we simply add up the values we calculated for each step. Therefore, the total amount of heat required to change 20.0g of H2O