Question

Calculate the heat, in kilocalories, that is absorbed if 183 g of ice at 0.0 ∘C is placed in an ice bag, melts, and warms to body temperature of 37.0 ∘C. The heat of fusion for water is 80. cal/g, or 334 J/g. The specific heat for water is 1.00 cal/g∘C, or 4.184 J/g∘C.

Answer 1

Answer:

The total amount of heat needed will be $Q_T=21.411kcal$.

Explanation:

We will divide the calculation in two: First, the heat needed to melt the ice, and then the heat needed to warm the resulting liquid from 0°C to 37°C.

$m=183g$

$l_f=80\frac{cal}{g} =334\frac{J}{g}$

$l_w=1\frac{cal}{g} =4.184\frac{J}{g}$

i) The fusion heat will be:

$Q_f=l_fm=14640cal=14.640kcal$

ii) The heat needed to warm the water from $T_i=0^{\circ}C$ to $T_i=37^{\circ}C$ will be:

$Q_w=l_wm(T_f-T_i)=6771cal=6.771kcal$

So, the total amount needed will be the sum of these two results:

$Q_T=Q_f+Q_w=14.640kcal+6.771kcal=21.411kcal$.