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θερμοτητα ασκησεις β γυμνασιου

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θερμοτητα ασκησεις β γυμνασιου

  1. 1. Example 1: How much heat (in calories) is needed to raise 20 grams of water from 5 o C to 40 o C? Q = m c (Tf—Ti) Q = x m = 20 gr c = 1 cal/go C (this is the specific heat of water) Tf = 40 o C Ti = 5 x = (20 g) (1 cal/go C)(400 C) -5 0 C) = 700 cal Example 2: How much heat (in calories) is needed to raise 140 grams of water from 20 o C to 25 o C? Q = mc(Tf—Ti) Q = x m = 140 grams c = 1 cal/go C (this is the specific heat of water) Tf = 25 o C Ti = 20 o C x = (140 g) (1 cal/g o C)(25-20 0 C) = 700 cal
  2. 2. Example 3: How much heat (in calories) is needed to raise 250 grams of water from 80 o C to 87 o C? Q = mc(Tf—Ti) Q = x m = 250 grams c = 1 cal/go C (this is the specific heat of water) Tf = 87 o CTi = 80 o C x = (250 g) (1 cal/goC)(87—80 0 C) = 1750 cal Example 4: How many calories are needed to raise50 grams of iron from55 oC to 200 oC? The specific heat capacity of iron is 0.11 cal/go C. Q = mc(Tf—Ti) Q = x m = 50 grams c = 0.11 cal/goC Tf = 200 o C Ti = 55 o C x = (50 g) (0.11 cal/go C)(200–55 0 C) = 797.5 cal
  3. 3. Example 5: How many grams of aluminum can be heated from90 o C to 120 o Cif 500 calories are applied? The specific heat of aluminum is 0.21 cal/go C. Q = mc(Tf—Ti) Q = 500 calories m = x grams c = 0.21 cal/go C Tf = 120 o C Ti = 90 o C 500 cal= (x) (0.21 cal/go C)(120—90 0 C) x = 79.4 grams Example 6: What is the specific heat capacity of a substanceif 400 calories cause 25 grams of it to go from 60 o Cto 190 o C? Q = mc(Tf—Ti) Q = 400 calories m = 25 grams c = x cal/go C Tf = 190 o C Ti = 60 o C 400 calories = (25 g) (x)(130 0 C) x = 0.123 cal/go C
  4. 4. Example 7: What is the final temperature if 500 calories are applied to 40 grams of copper at 20 o C? The specific heat capacity of copper is 0.092 cal/go C. Q = mc(Tf—Ti) Q = 500 m = 40 grams c = 0.092 cal/go C Tf = x Ti = 20 o C 500 cal = (40 g) (0.092 cal/goC)(x–20 0C) x = 156 o C Example 8: What was the initial temperature if 250 calories were applied to 100 grams of gold and the final temperature of the gold was 175 o C? The specific heat capacity of gold is 0.031 cal/go C. Q = mc(Tf—Ti) Q = 250 calories m = 100 grams c = 0.031 cal/go C Tf = 175 o C Ti = x o C 250 cal= (100 g) (0.031 cal/go C)(175—x0C) x = 94 o C

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