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Using the formula Q=mcΔT, the amount of heat energy required is (30g)(4.18J g-1 K-1)(12K) = 1504.8J = 360 calories.
Q=6*550*1.00q=3300
2pp
Specific heat for aluminium = 0.214 Heat required = 38.2 x 0.214 x (275 - 102) = 1414.24 calories
For water, the heat of fusion is 80 cal/g. So in other words, this is how many calories are needed to melt 1 gram of water that is frozen. Conversely, when you freeze 1 gram of water, you remove 80 calories of heat from it.... So, you multiply the calories needed to unfreeze a gram of water by the number of grams you have. In this case, 80 * 25 = 2000 Calories
103 calories. The heat of fusion of water is 80 cal/g and it takes one calorie to change the temperature of 1 g of water by 1 degree. 80+23=103 calories
It takes 1000 calories to heat 1 litre of water 1 degree C.
This heat is 51, 33 cal.
540
q (amt of heat) = mass * specific heat * temp. differenceThe specific heat of water is 1.00 cal/goC & the temperature difference is 70-30 = 40oCq = (105 grams)*(1.00 cal/goC)*(40oC)= 4,200 calories
1370 calories
there are 3000 calories in a water bottle so be careful!