Question

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?.

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. Express your answer to two significant figures and include the appropriate units.

B) Calculate the energy absorbed from a climber's body if he melts 0.45 kg of -15?C snow using a stove and drink the resulting 0.45 kg of water at 2?C, which his body has to warm to 37?C. Express your answer to two significant figures and include the appropriate units.

Answer 1

Part A

Mass of the ice = m = 0.45 kg

Initial temperature of ice = T₁ = -15 ^oC

Melting point of ice = T₂ = 0 ^oC

Final temperature of the water = T₃ = 37 ^oC

Specific heat of ice = C_i = 2100 J/(kg.^oC)

Specific heat of water = C_w = 4186 J/(kg.^oC)

Latent heat of fusion of water = L = 333 kJ/kg = 333000 J/kg

Heat absorbed from the climber's body = H

H = mC_i(T₂ - T₁) + mL + mC_w(T₃ - T₂)

H = (0.45)(2100)(0 - (-15)) + (0.45)(333000) + (0.45)(4186)(37 - 0)

H = 233721.9 J

Converting to significant figures,

H = 2.3 x 10⁵ J

Part B

Mass of the ice = m = 0.45 kg

Initial temperature of ice = T₁ = -15 ^oC

Temperature at which the water is taken in = T₂ = 2 ^oC

Final temperature of the water = T₃ = 37 ^oC

Specific heat of ice = C_i = 2100 J/(kg.^oC)

Specific heat of water = C_w = 4186 J/(kg.^oC)

Latent heat of fusion of water = L = 333 kJ/kg = 333000 J/kg

Heat absorbed from the climber's body = H

H = mC_w(T₃ - T₂)

H = (0.45)(4186)(37 - 2)

H = 65929.5 J

Converting to two significant figures,

H = 6.6 x 10⁴ J

A) Heat absorbed from the climber's body if he eats snow at -15 ^oC = 2.3 x 10⁵ J

B) Heat absorbed from the climber's body if he melts the snow and drinks it at 2 ^oC = 6.6 x 10⁴ J