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 1 year ago

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