± Heating and Cooling Curves Physical constants The constants for H2 are shown h

Question

± Heating and Cooling Curves Physical constants The constants for H2 are shown here: Consider heating solid water (ice) until it becomes liquid and then gas (steam). (Eiqure 1) Alternatively, consider the reverse process, cooling steam untl it becomes water and, finally, ice. (Figure 2) In each case, two types of transitions occur, those involving a temperature change with no change in phase (shown by the diagonal line segments on the graphs) and those at constant temperature with a change in phase (shown by horizontal line segments on the graphs). . Specific heat of ice: ce = 2.09 J/(g·°C) . Specific heat of liquid wte4.18 /(g C) . Enthalpy of fusion (H2O(s)H20(1)):A,"-334 J/g . Enthalpy of vaporization (H20(1)-H20(g): AHvap 2250 J/g Part A How much heat energy, in kilcjcules, is required to convert 62.0 g of ioe at-18.0 °C to water at 25.0 C? Express your answer to three significant figures and include the appropriate units. Hints 26.17 kJ Submit My Answers Give Up Incorrect; Try Again; 3 attempts remaining gure 1 of 2 Heating curve Part B How long would it take for 1.50 mol of water at 100.0 C to be converted completely into steam if heat were added at a constant rate of 24.0J/s? Express your answer to three significant figures and include the appropriate units Liquid and gas in equilbrium Hints 100 Solid and li n equilbrium Liquid Solid Value Units Heat added J) Submit My Answers Give Up

Explanation / Answer

PartA

Answer

27.8kJ

Explanation

There three heat requirements

Heat require to raise the tempersture of ice from -18 to 0,q1

q1 = m × T × Sice

m = mass of ice , 62g

T = tempereture difference , 0 - (-18) = 18

Sice = specif heat of ice , 2.09J/g

q1 = 62g ×18 × 2.09J/g

= 601.92J

Heat required to fuse ice,q2

Hfus = 334J/g

therefore,

q2 = 62g × 334J/g = 20708J

Heat required to raise the temperature of liquid water from 0 to 25 ,q3

q3 = m × T × Swater

   = 62g × (25 - 0) × 4.18J/g

   = 6479J

Therefore,

Total Heat energy required = 601.92J + 20708J + 6479J =27.8kJ

Part B

Answer

2.53×10^3s

Explanation

Enthalphy of Vaporization, Hvap = 2250J/g

No of mol of water = 1.50mol

mass of water = 1.50mol × 18g/mol = 27g

Heat required to Vapourize water, q = 27g ×2250J/g =60750J

Heating rate = 24J/s

Time required = 60750J/24(J/s) = 2531s

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