dang you chemistry!!!!!!?
ok so i finished all the mesasurement problems on this worksheet, then wtf
10. How many joules of heat are absorbed by 500.0g of water when its temperature increases from 20.0°C to 80.0°C? (sp. ht. water = 1.00cal/g°C)
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You will use the equation
Q = mass x specific heat x delta T = 500 g x 1.00 cal/g°C ( 80.0 - 20.0)= 30000 cal => 30.0 kcal
This is just a specific heat problem. The heat is related to the specific heat by the following formula:
heat = (mass)x(sp. ht.)x(change in T)
Replacing the values into this equation gives:
heat = (500.0 g)x(1.00 cal/(g*oC))x(80.0oC - 20.0oC) = (500.0 g)x(1.00 cal/(g*oC))x(60.0oC) = 30000 cal or 30.0 kcal
mcÎT = Q: m = 500.0g; c = 1.0cal/g/°C; ÎT = 80°C - 20 = 60°C. 1.0cal = 4.184 Joules.
500g x 1.0cal/g/°C x 60°C x 4.184J/cal = 125,520 J (= 125.5kJ).
(Check using kg and kJ/kg/°C:
0.500kg x 4.184kJ/kg/°C x 60°C ÎT = 125.5kJ)