How many calories are needed to melt 80.0 g of ice at 0 degrees celcius and to warm the liquid to 65 degrees celcius(two steps)??? I really need help with this question. i'm so confused.
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Hi Sarah!
This problem deals with calorimetry and believe me, this stuff is confusing lol. So there is indeed two steps for this problem. The first step is for us to find the amount of heat to melt the ice and the second step is for us to find the amount of heat needed to heat the water (now a liquid) to 65 degrees celsius. Always remember that it actually takes extra heat to melt water because many people forget that. So to find how much heat is needed to heat the water, you need to look up the heat of fusion for water which is 79.72 cal/g. This value just means that for every gram of ice, it takes about 80 calories of heat to heat it up to a liquid. We have to melt 80 grams of ice so use dimensional analysis and find the amount of heat we need: (80g)(79.72cal/1g) = 6377.6 calories needed to melt all that ice. Now we're done with the first step and for the second step, we need to use the calorimetry equation which is q = mc(delta T) where m represents the mass of the substance, c represents the specific heat of the substance, delta T represents the change of the temperature and q represents the amount of heat required in joules. Now, we can plug in all the variables we know. Also be aware to look up the specific heat for water ( which is 4.184 J/G X C). Now we have q = (80g)(4.184 J/GxC)(65-0). For the temperature, the value is always the final temperature minus the initial temperature. The question says that the water is heated to 65 degrees, so that is the final, and we use 0 degrees as the initial because we are assuming that the ice has just melted to a liquid, Solve for q which is about 21,756.8 joules. Now we want our answer in calories so use the conversion factor: 1 calorie = 4.148 joules to change our units. (21,756.8 joules)(1 calorie/ 4.148 joules) = about 5245 calories. Now the total amount of heat is then the heat needed in the first step plus the heat needed in the second step (5245 + 6378) = 11,622 calories needed. This is answer is not in significant figures if that matters at the moment.
Good luck on chemistry and I hope this wasn't too confusing. Feel free to ask me any more questions!! :)
This problem deals with calorimetry and believe me, this stuff is confusing lol. So there is indeed two steps for this problem. The first step is for us to find the amount of heat to melt the ice and the second step is for us to find the amount of heat needed to heat the water (now a liquid) to 65 degrees celsius. Always remember that it actually takes extra heat to melt water because many people forget that. So to find how much heat is needed to heat the water, you need to look up the heat of fusion for water which is 79.72 cal/g. This value just means that for every gram of ice, it takes about 80 calories of heat to heat it up to a liquid. We have to melt 80 grams of ice so use dimensional analysis and find the amount of heat we need: (80g)(79.72cal/1g) = 6377.6 calories needed to melt all that ice. Now we're done with the first step and for the second step, we need to use the calorimetry equation which is q = mc(delta T) where m represents the mass of the substance, c represents the specific heat of the substance, delta T represents the change of the temperature and q represents the amount of heat required in joules. Now, we can plug in all the variables we know. Also be aware to look up the specific heat for water ( which is 4.184 J/G X C). Now we have q = (80g)(4.184 J/GxC)(65-0). For the temperature, the value is always the final temperature minus the initial temperature. The question says that the water is heated to 65 degrees, so that is the final, and we use 0 degrees as the initial because we are assuming that the ice has just melted to a liquid, Solve for q which is about 21,756.8 joules. Now we want our answer in calories so use the conversion factor: 1 calorie = 4.148 joules to change our units. (21,756.8 joules)(1 calorie/ 4.148 joules) = about 5245 calories. Now the total amount of heat is then the heat needed in the first step plus the heat needed in the second step (5245 + 6378) = 11,622 calories needed. This is answer is not in significant figures if that matters at the moment.
Good luck on chemistry and I hope this wasn't too confusing. Feel free to ask me any more questions!! :)
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Sarah, First write the two steps.
Step 1: 80.0 g H2O(s) @ 0°C --> 80.0 g H2O(l) @ 0°C
Use the formula: H(1) = mLf, where m = mass in grams and Lf = latent heat of fusion of ice = 334 J/g (or 334 kJ/kg). H(1) = (80.0 g)(334 J/g)
Step 2: 80.0 g H2O(l) @ 0°C --> 80.0 g H2O(l) @ 65°C
Use the formula: H(2) = mcΔT, where c = specific heat of water = 4.184 J/g°C, and ΔT = T(final) - T(initial) H(2) = (80.0 g)(4.184 J/g°C)(65°C - 0°C)
Total heat required = H(T) = H(1) + H(2)
Hope this is helpful to you. JIL HIR
Step 1: 80.0 g H2O(s) @ 0°C --> 80.0 g H2O(l) @ 0°C
Use the formula: H(1) = mLf, where m = mass in grams and Lf = latent heat of fusion of ice = 334 J/g (or 334 kJ/kg). H(1) = (80.0 g)(334 J/g)
Step 2: 80.0 g H2O(l) @ 0°C --> 80.0 g H2O(l) @ 65°C
Use the formula: H(2) = mcΔT, where c = specific heat of water = 4.184 J/g°C, and ΔT = T(final) - T(initial) H(2) = (80.0 g)(4.184 J/g°C)(65°C - 0°C)
Total heat required = H(T) = H(1) + H(2)
Hope this is helpful to you. JIL HIR