Calculate the heat needed at 0∘C∘to make each of the following changes of state. Calories to melt 22 g of ice

According to the heat equations: Q = C1mDelta;T1 + C2mDelta;T2 + lambda;m. m = 0.2 kg For water: C1 = 4.183x103 [J / (kg K)] Delta;T1 = 25 - 0...

Let’s first consider the processes that occurred with the initial ice. Firstly, the ice was heated to 0°C (melting point of water). Then, the part of ice melted to form...

Enthalpy of vaporization for water is 2260 kJ/kg. Thus, the amount of RELEASED heat will be Q = r*m = 2260 kJ/kg*0.010 kg = 22.6 kJ or 5.4 kcal.

Mollar mass (C6H6) = 78 g/mol n = 7.87 / 78 = 0.101 (mol) Q = 0.101 * 9.92 = 1.002 (kJ)

Find the total heat used by ice in changing state from solid to liquid(water) Q1 = 20g*80cal/g = 1600 cal 1600 cal = 6694.4 J = 6.69 kJ M'C(T'-x) =...

Q=mc∆T.Q= 9.8 x 2.1 x 10 = 205.8 J.

80 calories needed to melt 1gm ICE at 0 degree.Hence for 90 gm ICE, it require 80x90 = 7200 calories So answer = 7200 calories.

Heat of water vaporization 2260 kJ/kg. 1 kJ = 0.239 kcal, 2260 kJ/kg = 540.14 kcalQ = 540.14*0.155 = 83.7217 kcal = 83721.7 cal.

Solution :Sublimation is the transition of a substance directly from the solid to the gas phase without passing through the intermediate liquid phase.Dry ice sublimes easily. Iodine will sublimes easily...

5.2°C = (273 + 5.2)k = 278.2kChange in temperature = 368k -278.2= 89.8kq (heat energy) = MCp∆TWhen Cp of water =1046J q = 7.59 g × 1046J × 89.8k =...