He specific heat of water is 4.18 j/(g * °c) how much heat, in kilojoules, must be added to 250 g of water to increase the temperature of the water by 5.0°c? report your answer with 3 significant figures.
Q=McDeltaT Q= 250g×4.18j/g°c×5.0°c Q=5225 joules To get the kilojoules now I will have to divide the joules by 1000. [tex]q = \frac{5225}{1000} = 5.225kj[/tex]
