1). Write the equation used to determine the latent heat of fusion of ice in ter
ID: 1912152 • Letter: 1
Question
1). Write the equation used to determine the latent heat of fusion of ice in terms of the appropriate masses, temperature changes and specific heats (define all quantities) in an adiabatic process. Check the experiment for the items making up the system. Have the ice start off at 0?C. Keep the equation that you are writing in non-numerical terms. Starting at 0?C is telling you that you will have a term involving the latent heat of fusion and another term of warming to the equilibrium temperature. State in one sentence the basic principle of physics expressed by this equation. 2). How much heat is absorbed by 100 grams of water going from 0Explanation / Answer
1. How much heat is absorbed by 1.00 x 10^2 g of ice at -20.0 deg C to become water at 0 deg C ? Stage1:- 1.00 * 10^2 g of ice's temperature changes from -20 deg C to 0 deg C Specific heat capacity of ice = 2.05 J/g/deg C Heat abosorbed = 1.00 * 10^2 * 2.05 * [0 - (-20)] = 4,100 J Stage 2:- 1.00 * 10^2 g of ice becomes water Latent heat of melting of ice = 333.55 J/g Heat absorbed = 1.00 * 10^2 * 333.55 = 333,55 J Total heat absorbed = 4,100 J + 33,355 J = 37,455 J 2. A 2.00 x 10^2 g sample of water at 60 deg C is heated to steam at 140 deg C. How much heat is absorbed? Stage 1:- 2.00 x 10^2 g of water's temp changes from 60 deg C to 100 deg C Specific heat capacity of water = 4.18 J/g/deg C Heat absorbed = 2.00 * 10^2 * 4.18 * (100 - 60) = 33,440 J Stage 2:- 2.00 * 10^2 g of water becomes steam Latent heat of vaporization of water = 2256 J/g Heat absorbed = 2.00 * 10^2 * 2256 = 451,200 J Stage 3: 2.00 * 10^2 g of steam's temp. changes from 100 deg C to 140 deg C Specific heat capacity of steam = 2.08 J/g/deg C Heat absorbed = 2.00 * 10^2 * 2.08 * (140 - 100) = 16,640 J Total heat absorbed = 33,440 J + 451,200 J + 16,640 J = 501,240 J 3. How much heat is needed to change 3 x 10^2 g of ice at -30 deg C to steam 130 deg C? Can you solve this yourself? There are 5 stages. First: Ice's temp. changes from - 30 deg C to 0 deg C. Second: Ice changes into water. Third: Water's temp. changes from 0 deg C to 100 deg C. Fourth: Water becomes steam. Fifth: Steam's temp. changes from 100 deg C to 130 deg C Find heat for each stage and then add them. 4. A 2 x 10^4 sample of water at 80 deg C is mixed with 2 x 10^4 g of water at 10 deg C . Assume no heat loss to the surroundings. What is the final temperature of the mixture? Both samples of water are of equal amount. So final temp. will be in the middle. Final temp. = average of 80 deg C + 10 deg C = 45 deg C What if they are not in equal proportion? For that, I will solve the same problem in another way, which you can use whether samples are equal or not. Let c = specific heat capacity of water t deg C = final temp. For first sample, temp. decreases from 80 deg to t deg. Heat lost by the sample = 2 * 10^4 * c * (80 - t) For second sample, temp. increases from 10 deg to t deg. Heat gained by sample = 2 * 10^4 * c * (t - 10) Heat lost = heat gained 2 * 10^4 * c * (80 - t) = 2 * 10^4 * c * (t - 10) Or 80 - t = t - 10 Or 2t = 90 Or t = 45 5. A 4.00 x 10^2 g sample of methanol at 16 deg C is mixed with 4 x 10^2 g of water at 85 deg C. Assume no heat loss to the heat surroundings. What is the final temperature of the mixture? Can you solve this yourself? Final temp. will be somewhere between 16 deg and 85 deg Let final temp = t deg C Methanol gains heat because its temp increases. Initial temp of methanol is 16 deg and final is t deg. Use formula mass * specific heat capacity * change in temp. to get heat gained. Change in temp = t - 16 Water loses heat because its temp. decreases. Initial temp is 85 deg and final is t. Use formula mass * specific heat capacity * change in temp. to get heat lost. Change in temp = 85 - t Then heat lost = heat gained. You will need specific heat capacities of methanol and water. If temp. goes up, then the sample gains heat. If goes down, then the sample loses heat. For change in temp., use higher temp minus smaller temp. 6. How much heat is absorbed by 60 g of copper when it is heated from 20 deg C to 80 deg C ? Mass = 60 g Change in temp. = 80 - 20 = 60 deg Use formula mass * specific heat capacity * change in temp. You will need specific heat capacity of copper. 7. A 38 kg block of lead is heated from -26 deg C to 180 deg C. How much heat does it absorb during the heating? Mass = 38 g Change in temp. = 180 - (-26) = 206 deg Use formula mass * specific heat capacity * change in temp. You will need specific heat capacity of lead.
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