Question

What volume of methane at STP would be required to convert a 2.5 kg block of ice at −13 ∘C to water at 13 ∘C, assuming that no heat is lost? The heat of fusion for ice is 6.01 kJmol−1 . The heat capacity of ice is 2.108 Jg−1∘C−1, the specific heat capacity of water is 4.184 Jg−1∘C−1 and the heat of combustion for methane is 891 kJmol−1.

Answer 1

Mass of ice. = 2.5 kg = 2500g now, total energy required by ice = energy required to increase the temp of ice from -13 degree c to 0 degree c + energy required for few ions + energy required to increase the temp. of water from 0 degree c to 1322 degree c now, energy required to increase the temp of ice (from-13 degree c to 0 degree c0 = mass times heat capacity times delta T = [2500g times 2.108 times 90-(13)]J = 68510J energy required for fusion of ice = (25009/189 mol times 6.01 times 10^3 J mol^-1 834722.2 J energy required to increase the temp of water them 0 degree c to 13 degree c = [2500 g times 40184 times (13-0)]J = 135980J therefore Total energy required = (68570 + 834722.2 + 135980)J = 1039212.2J = 1039.21KJ now heat of combustion of methane = 891 KJ mol^-1 therefore no. of moles of methane required = 1039.21/891 mol = 1.166 mol At STP 1 mol of gas occupies 22.4L