Question

Cold-winter survival experts do not recommend eating snow for hydration. One reason for this is the energy it takes to increase temperature of the snow, melt it, and then increase the temperature of the resulting water to body temperature. Calculate the total amount of heat to convert 2.000 kg of snow at -15.00 0C to liquid water at 38.00 0C.
POTENTIALLY HELPFUL INFORMATION:
The heat of fusion for water is 1.44 kcal/mol.
The specific heat capacity of solid water is 0.488 cal/g0C.
The specific heat capacity of liquid water is 1.00 cal/g0C.
The freezing point of water is 00C (also known as melting point).

Answer 1

Answer:

The total amount of heat is 250.64 kcal.

Explanation:

Given that,

Total amount of heat = 2.000 kg

Temperature of snow = -15.00°C

Temperature of water = 38.00°C

Heat of fusion of water = 1.44 kcal/mol

Specific heat of solid water = 0.488 cal/g°C

Specific heat of liquid = 1.00 cal/g°C

We need to calculate the heat required to bring temperature -15°C to 0°C snow

Using formula of heat

$Q_{1}=mc_{p}\Delta T$

Put the value into the formula

$Q_{1}=2000\times0.488\times(0+15)$

$Q_{1}=14640\ cal$

$Q_{1}=14.640\ kcal$

We need to calculate the heat required to convert 0°C snow  into 0°C water

Using formula of heat

$Q_{2}=ml$

Put the value into the formula

$Q_{2}=\dfrac{2000}{18}\times1.44$

$Q_{2}=160\ kcal$

We need to calculate  heat required to increases temperature of water from 0°C  to 38°C

Using formula of heat

$Q_{3}=mc_{p}\Delta T$

Put the value into the formula

$Q_{3}=2000\times1.00\times(38-0)$

$Q_{3}=76000\ cal$

$Q_{3}=76\ kcal$

We need to calculate the total heat

Using all heat amount

$Q=Q_{1}+Q_{2}+Q_{3}$

Put the value into the formula

$Q=14.640+160+76$

$Q=250.64\ kcal$

Hence, The total amount of heat is 250.64 kcal.