How much heat is needed to change 7.50 g of silver from solid at 25C to liquid at 961C? (specific heat of silver is 0.237 J/g-C, melting point is 961C, heat of fusion is 11.0 J/g).
This is what I've done...
(7.50g)(0.237 J/g-C)(936C) = 1663.745J
(11.0J/g)(7.50g) = 82.5 J
total heat required = 1663.74 + 82.5 = 1746.24
Is that right???
HPV
Favorite Answer
Yes it is.
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first of all you're coping with good to liquid to gasoline alterations. next, utilizing good ice as an party, going to liquid water, then on to water vapor. purely some energy or joules had to warmth each and each gram of ice one degree C Then once you attain the melting ;element of ice, about 80 energy warmth power had to soften each and each gram of ice at 0 C to one gram of water at 0 ranges C (so there's a large number of power enthusiastic about going from one area to the subsequent) The it takes about one calorie or 4.184 joules of power to warmth one gram of water one degree C each and each of ways from 0 ranges to l00 ranges C. Then, an excellent number of power is significant to flow that one gram of water at l00 ranges C to one grams of water vapor at l00 ranges C. This power is termed "the nice and cozy temperature of vaporization". It quantities to about 540 energy in conserving with gram of water. This power is used to split liquid water molecules into vapor. yet another thrilling element about area alterations is that the temperature stages off at the same time as the replace from one area to a unique is happening. party, once water reaches its boiling element at one surroundings stress, its temperature holds at l00 ranges C till each and each of the water has vaporized.