Question

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.

Answer 1

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

Where,

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

Now,

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

                                                = 10800 cal + 4995 cal

                                                = 15796 cal

                                                = 15.80 kcal