The so called \"ice-weight reduction diet\" has been considered as a method for

Question

The so called "ice-weight reduction diet" has been considered as a method for counterweight gain due to the intake of excessive calories. The weight-gains counteraction results from the facts that: H2O(s) consumes heat as it is melted and cold H2O(l) consumes heat as it is warmed to normal body T. Since adult humans consume an average of 2500 calories a day, calculate the quantity of ice at 273K which must be "eaten" to cancel the effect of 1000 cal food intake.

Note: For H2O(s) at 273K, deltaH(fusion) = 6.01 kJ/mol. Normal Body T = 37.0 degrees C. Cal is a "large" or "food" calorie where: 1 cal = 1kcal = 4.184 kJ

See if your calculations suggest that this particular diet is a reasonable way to combat weight gain

Please show step by step work so I can learn

Explanation / Answer

(4.184 kJ/6.01 kJ/mol)=0.696 mol

H2O=18g/mol

18g/mol*0.696 mol=12.53g

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