Question

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A hungry student placed a pot of spaghetti water on a hot plate which had power 1.0 kW. The pot contained 2.01 water with temperature 11 °C. The following values apply to water; thermal capacity 4.18 JK-1g-1, density 1.0 g/ml, evaporation enthalpy 41 kJmol-1. (a) Calculate the temperature of the water after the hob has been running for 10 my. Overlook the heating of the saucepan and stove top. (b) Calculate the temperature of the water after the hob has been started another 5 min, ie a total of 15 min the answers are a) 83 °C b) 100 °C

Answer 1

1. The hot plate Supplies 1000 J/S heat. ... In 10 mre's the total heat supplied by the hot plate = 1000x10x6. Meat suppoea y = 6*100J The mass of water = (2x1000x1) 9 = 2000 9| ensily = 1 g/me] specific heat of water = 4.18 J. Kg Initial temperature of water = lle a Let the final temperature of water = tc . · Total heat absorbed by water = 2000 x4.18 *(t-11) J : 2000X4:18x(+-11) = 68100 or (t-11) = - or } = (7.77 +11) = 82.77–83 ... The teinperature of the water will be 85ć after 10 min. 2. Total heat subblied in 15 min = 15x60x1000 J =9x10 J Now, let the final temperature be tice * Heat heat absorbed by water = 2000 x4.18 (-11) J 2000 x 4.18 = 71.77

.: 2000*4.1$(t-11) = 94100 or fl-" = 98105. ce = 107.65 2000x4'18 or ti =(107.65 +11) = 118.65 The vatue of ti indicates that there will be a phase change (liquid to vapor) for water.... from uc to . Now, heat required to raise the temp, of water 1000 = 2000x4:18*(100-11). J = 7,44*10'J Heat required to vaporise 2000 g water = 2000 x41*1000J . Total heat required = 7:44+ 45.56) x 105 j = 53.00 x 10 J -0,5 45.56 *10° Jan But total heat supplied by hot plate = 9x105J. These indicates that some amount of water will be reporised at boiling point of water (1006) and boiling will be continewing, so the final temperature will be 1000 18