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:

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

Answer 2

The movement of energy from a hot to a cold item is characterized as heat. Heat energy flows from a hot material to a cold one. The amount of heat absorbed will be 21.411 kcal.

What is heat?

The movement of energy from a hot to a cold item is characterized as heat. Heat energy flows from a hot material to a cold one.

This occurs because faster-vibrating molecules transmit their energy to slower-vibrating ones. Heat content is another name for vibrational energy.

The following data are given in the problem

C IS specific heat for water = 1.00 cal/g∘C or 4.184 J/g∘C.

$\rm L_f$ is the heat of fusion of ice = 80. cal/g, or 334 J/g.

$\rm L_w$ is the heat of fusion for water is 80. cal/g, or 334 J/g

$\rm C$ is the specific heat for water =1.00 cal/g∘C or 4.184 J/g∘C.

In order to warm the body up to 37°C.First, ice is melted due to which latent heat if fusion comes into the picture. After that, the sensible heat is observed from 0 to 30°C.

The heat required to melt the ice;

$\rm{Q_f=l_fm}\\\\\rm{Q_f= 334 \times 183}\\\\\rm{Q_f=14640 cal=14.6401 Kcal.$

Sensible heat will be

$\rm Q_W= L_wm(t_f-t_i)\\\\\rm Q_W= 334 \times183(37^0-0^0))\\\\\rm Q_W=6771 Kcal=6.771Kcal.$

The total amount of heat

$\rm{Q_t= Q_f+Q_w}\\\\\ \rm{Q_t= 14.06+6.771}\\\\\rm{Q_t=21.441\; Kcal$

Hence the amount of heat absorbed will be 21.411 kcal.

To learn more about the heat refer to the link;

https://brainly.com/question/1429452