Answered Essay: A sports trainer applies an ice bag to the back of an injured athlete. Calculate the heat in kcal

A sports trainer applies an ice bag to the back of an injured athlete. Calculate the heat in kcal that is absorbed if 135 g of ice at 0 Celsius is placed in an ice bag, melts, and rises to body temperature at 37 Celsius. (For water, 80 cal(334 J) is needed to melt 1 g of ice or must be removed to freeze 1 g of water.) express the heat to 2 significant #’s and include appropriate units.

Expert Answer


Ans. For ice at 00C to reach 370C, it must absorb heat during following two distinct processes-

A. Fusion of ice: The solid ice melts into liquid water at 00C. Note that the temperature remains constant at melting point of ice.

B. Liquid water at 00C absorbs heat and raises its temperature to 370C.

So, total amount of heat to be absorbed by ice can be calculated in following two steps-

Step 1: Heat absorbed to convert solid ice into liquid at 0.00C is given by-

q1 = m x C     – equation 1

where, m = mass of ice        ; C = heat capacity of ice

or, q1 = 135 g x (80 cal / g)

hence, q1 = 10800 cal

Step 2: Heat required to warm liquid at 0.00C to 370C is given by-

q = m s dT                            – equation 2


q = heat absorbed

m = amount of water

s = specific heat of water [ 1 cal g-10C-1 ]

dT = Final temperature – Initial temperature = (T2 – T1)0C

= 37.00C – (0.00C) = 370C

Putting the values in equation 2-

q = 135 g x (1 cal g-10C-1) x 370C = 4995 cal


Total amount of heat absorbed during the process = q1 + q

= 10800 cal + 4995 cal

= 15796 cal

15.80 kcal

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