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Water has a specific heat of 4.184 j/g-C. if 300.0 grams of water absorb 2.50 x 10 J of energy, how much will the temperatur
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Answer #1

tho formula is

q=mc47

q = 2.50 x 104 J

c= 4.184 J/g.oC

mass = m = 300 g

We are asked to find delta T

2.50 x 10 = 300 x 4.184 Χ ΔΤ

AT = 2.50 x 104 300 x 4.184

AT = 19.9°C = 20°C

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