Question

1)

High-altitude mountain climbers do not eat snow, but always melt it first with a stove. To see why, calculate the energy absorbed from a climber's body under the following conditions. The specific heat of ice is 2100 J/kg⋅C∘, the latent heat of fusion is 333 kJ/kg, the specific heat of water is 4186 J/kg⋅C∘. Part A Calculate the energy absorbed from a climber's body if he eats 0.45 kg of -15∘C snow which his body warms to body temperature of 37∘C.

2)

An insulated beaker with negligible mass contains liquid water with a mass of 0.275 kgkg and a temperature of 61.1 ∘C∘C .

Part A How much ice at a temperature of -14.3 ∘C∘C must be dropped into the water so that the final temperature of the system will be 33.0 ∘C∘C ? Take the specific heat of liquid water to be 4190 J/kg⋅KJ/kg⋅K , the specific heat of ice to be 2100 J/kg⋅KJ/kg⋅K , and the heat of fusion for water to be 3.34×105 J/kgJ/kg .

Answer 1

1.

Part A.

Energy absorbed when climber eats the snow without melting it

Q = Q1 + Q2 + Q3

Q1 = Energy required to change -15 C ice to 0 C ice = mi*Ci*dT1

Q2 = energy required for phase change = mi*Lf

Q3 = energy required from 0 C water to 37 C water = mi*Cw*dT2

Q = mi*Ci*dT + mi*Lf + mi*Cw*dT

Given values are:

mi = mass of ice = 0.45 kg

dT1 = 0 - (-15) = 15 C

dT2 = 37 - 0 = 37 C

Using these values:

Q = 0.45*2100*15 + 0.45*333*10^3 + 0.45*4186*37

Q = 233721.9

Q = 2.34*10^5 J = energy absorbed by eating snow

2.

Suppose mass of ice required is mi, then Using energy conservation:

Energy absorbed by ice = energy released by water

Q1 + Q2 + Q3 = Q4

Q1 = Energy absorbed by ice from -14.3 C to final temperature 0 C = mi*Ci*dT1

Q2 = Energy absorbed by ice during phase change from ice to water = mi*Lf

Q3 = Energy absorbed by water from 0 C to final temperature 33.0 C = mi*Cw*dT2

Q4 = Energy released by water from 61.1 C to final temperature 33.0 C = mw*Cw*dT3

Given values are:

mw = mass of water = 0.275 kg

mi = mass of ice = ? kg

dT1 = 0 - (-14.3) = 14.3 C

dT2 = 33.0 - 0 = 33.0 C

dT3 = 61.1 - 33.0 = 28.1 C

Using these values:

mi*Ci*dT1 + mi*Lf + mi*Cw*dT2 = mw*Cw*dT3

mi = (mw*Cw*dT3)/(Ci*dT1 + Lf + Cw*dT2)

mi = (0.275*4190*28.1)/(2100*14.3 + 3.34*10^5 + 4190*33.0)

mi = 0.0645 kg

Mass of ice required = 0.0645 kg

Let me know if you've any query.