Question

A sports trainer applies an ice bag to the back of an injured athlete. Calculate the heat in kcal that is absorbed if 145 g of ice at 0.0 ∘C is placed in an ice bag, melts, and rises to body temperature of 37.0 ∘C. (For water, 80. cal (334 J) is needed to melt 1 g of ice or must be removed to freeze 1 g of water.)

Answer 1

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mass of bice = 145 gms => Latent heat of fusion = 80 callam => O = Orlice wate8) + 0° 0°c - ML +ms at Os (wate8 0°C water ) 37°c => (145 x 80 cal) + (1458 IX37) =) (145) (80 +37) = 169.65 kcal