hw: q 10
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hw: q 10
I need help. For this question, they are asking: An ice cube with a mass of 54.2 g at 0.0 ∘C is added to a glass containing 449 g of water at 45.0 ∘C . Determine the final temperature of the system at equilibrium. The specific heat capacity of water, Cs, is 4.184 J/g⋅∘C , and the standard enthalpy of fusion, ΔH∘fus , of water is 6.01×103 J/mol. Assume that no energy is transferred to or from the surroundings ". when I tried to tackle this problem I ended up getting 33 C which was wrong. how can I approach this problem?
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- Posts: 47
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Re: hw: q 10
For this problem, you have to consider the heat of fusion necessary to melt the ice as well as the heat necessary to bring it up to the temperature of the water. However, the water will also lose heat, so the final temperature should be lower than 45 C. To set up this problem, you need to assume that all of the heat that flows out of the water flows into the ice cube with no heat lost (q of the ice = -q of the water). So the equation would be (moles of ice)(ΔHfus) + (mass of ice)(Cs)(T - 0) = -[(mass of water)(Cs)(T - 45)].
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