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At 1 atm, how much energy is required to heat 55.0g of H2O(s) at -22.0 ^degree...

At 1 atm, how much energy is required to heat 55.0g of H2O(s) at -22.0 ^degree C to H2O(g) at 169.0^degree C?

Solutions

Expert Solution

q1   = mct

       = 55*2.087*(0-(-22)

       = 55*2.087*22   = 2525.27J

q2   = m * heat of fustion

       = 55*333.6 = 18348J

q3    = mct

         = 55*4.184*(100-0)

        = 23012J

q4    = m*heat of vaporisation

          = 55*2257

         = 124135J

q5     = mct

        = 55*2*(169-100)

       = 55*2*69

        = 7590J

q =   q1+ q2+q3+q4+q5

= 2525.27+18248+23012+124135+7590    = 175510J   = 175.51KJ


queen_honey_blossom answered 2 years ago
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