calculate the amount of J required to heat 115.0g of solid H2O at -25 degrees C...

calculate the amount of J required to heat 115.0g of solid H2O at -25 degrees C to liquid H20 at 30 degrees C

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Expert Solution

-25 ^oC ------------------> 0 ^oC----------------------> 30^oC

(1) (2) (3)

there are 3 energy conversions

Q = Q1 + Q2 + Q3

Q1 = m Cp dT

Q1 =115 x 2.09 x 25

Q1 = 6009 J

moles of soldi = 115 / 18 = 6.39

Q2 = n x delta H fusion = 6.39 x 6.01 x 10^3

= 38397 J

Q3 = m Cp dT

Q3 = 115 x 4.184 x 30

Q3 = 14435 J

Q = Q1 + Q2 + Q3

Q = 58841 J

amount of heat required = 58841 J

queen_honey_blossom answered 1 year ago

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