Chem Help -- Specific Heat Capacity?

Calculate q (in kJ) when 287 g of water is heated from 29.8°C to 87.5°C. The specific heat capacity of water is 4.184 J/g.°C. Express your answer in scientific format: 1.25 x 103 = 1.25E3

Chem Help -- Specific Heat Capacity?
The total heat energy Q = m * Cp * ΔT





Here Q = heat required to raise the temperature from T1 to T2


m = mass of water in kg (287/1000 = 0.287 kg)


Cp = Specific heat of water (4.184 KJ/kg C)


ΔT = Temperature difference (here 87.5 - 29.8)





Substituting





Q = 0.287 * 4.184 * (87.5 - 29.8) = 69.28 KJ (kilojoule) = 0.6928 * 10^2 KJ or 0.6928E2





It can also be written as 6.928 * 10^4 Joule. 6.928 E4 Joule.
Reply:Specific heat capacity formula is q=(mass)(specific heat) (change in temp).





Change in Temperature= 87.5-29.8 = 57.7 deg C





q = (287 g H20)(4.18 J/(g*deg C))(57.7 deg C) = 6.92E4 Joules which equals 6.92E1 kJ.





Hope this helps!
Reply:ur book should have the formula its like delta C or something i forget exactly what is.

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