Chapter 14, Problem 28P

To determine

(a) To Determine:

Energy absorbed by the body when we eat 1 kg of snow at $-{\text{15}}^0\text{C}$.

Answer to Problem 28P

Solution:

$\text{519 kJ}$

Explanation of Solution

Given info:

Heating from $-{\text{15}}^0{\text{C to 37}}^0\text{C}$

(body temperature)require temperature change of $\Delta T={\text{52}}^0\text{C}$.

$\begin{array}{l}\text{Latent heat of fusion}=\text{333 kJ}/\text{kg},\\ \text{specific heat of water}=\text{4186 J}/{\text{kg}}^0\text{C}\\ \text{Specific heat of ice is 21}00\text{ J}/{\text{kg}}^0\text{C} \end{array}$

Formula used:

The heat energy absorbed or released to raise the temperature by one degree is given by;

$\begin{array}{l}Q=mc\Delta T\\ Q=mL\end{array}$

Here,

m = mass of the substance.

C = specific heat.

◻T= Temperature difference.

L=latent heat of fusion.

Calculation:

As there is phase change at 0⁰C, the energy needed is given by;

$\begin{array}{l}E=mc\Delta T\\ E={{\displaystyle m}}_i\times {{\displaystyle c}}_i\times \left(0-(-{15}^0)\right)+{{\displaystyle m}}_1\times {{\displaystyle H}}_i+{{\displaystyle m}}_w\times {{\displaystyle c}}_w\times \left(0-(-{37}^0)\right)\\ E=1\text{ kg}\times 2100\text{ J/kg}\times \text{(}{{\displaystyle 15}}^0\text{C)+1kg}\times \text{333000}\text{J/kg +1 kg}\times (\text{4186 J/kg)}\times {{\displaystyle 37}}^{0}C\\ E\text{=519 kJ}\end{array}$

To determine

(b) To Determine:

Energy absorbed by the body when we drink melted 1 kg of -15⁰ C snow and convert it into 1 kg water at 2⁰ C.

Answer to Problem 28P

Solution:

$\text{131}\text{.9 kJ}$

Explanation of Solution

Calculation:

When water is at 2⁰ C and climbers drink it then energy required by body to absorb It is given by;

$\begin{array}{l}E=mc\Delta T\\ E={{\displaystyle m}}_i\times {{\displaystyle c}}_i\times \left({{\displaystyle 37}}^{0}C-(2^0)\right)\\ E\text{=131}\text{.9 kJ}\end{array}$