48g of ice at its melting point is heated with steam at its condensation point. The final temperature of the system is 50 degree Celsius. How much steam was used to melt the ice?
q=(48g)(1mol/18g)(50-0C)(0.97kcal/mol*C) = 129kcal
Not sure where to go from here (or if that first part is even right)?
To melt the ice you have
48*heat fusion = ?
To raise T from zero to 50 C you have
48 x specific heat H2O x (50-0) = ?
Total you need to melt the ice and raise it to 50 C is ? + ?
You get x grams steam x heat condensation = ?
You get x grams H2O from steam x specific heat H2O x (100-50) = ?
The total you get from steam is ?x + ?x
Set them equal to each other and solve for x = grams steam. I did a quickie calculation and came up with about 10 or 11 grams.
I understand up to the part about x grams of steam. Is x grams water from steam something I'm supposed to know or no? Also why am I totaling up ?x +?x, but then setting them equal to each other...? Can you explain that last part in more details, please? Thank you so much!!
To solve this problem, let's break it down step by step.
Step 1: Determine the heat energy required to melt the ice.
The heat energy required to melt a solid substance is given by the formula:
q = m * ΔHfus
where q is the heat energy in calories, m is the mass of the substance in grams, and ΔHfus is the heat of fusion (also known as the latent heat of fusion) of the substance, which represents the amount of heat energy required to change the substance from a solid to a liquid at its melting point.
For ice, the heat of fusion is 1 calorie per gram. Therefore, we can calculate the heat energy required to melt the ice as follows:
q = (48g) * (1 cal/g) = 48 calories
Step 2: Determine the amount of steam used to provide this heat energy.
To calculate the amount of steam used, we need to use the equation:
q = m * c * ΔT
where q is the heat energy in calories, m is the mass of the substance in grams, c is the specific heat capacity of the substance (the amount of heat energy required to raise the temperature of 1 gram of the substance by 1 degree Celsius), and ΔT is the change in temperature.
For steam, the specific heat capacity is 0.97 cal/(g * °C). We know the change in temperature is 50 °C (as the final temperature of the system is 50 °C).
Substituting the known values into the equation, we can solve for the mass of steam used:
48 calories = m * (0.97 cal/(g * °C)) * 50 °C
m = (48 calories) / ((0.97 cal/(g * °C)) * 50 °C)
m ≈ 0.98 g
Therefore, approximately 0.98 grams of steam were used to melt the ice.
Note: The initial calculation where you converted the mass of the ice to moles and multiplied it by the change in temperature and the heat of fusion is not necessary in this case, as the problem directly asks for the mass of steam used.