High-altitude mountain climbers do not eat snow, but always melt it first with a

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/kgC, the latent heat of fusion is 333 kJ/kg, the specific heat of water is4186 J/kgC.

Part A

Calculate the energy absorbed from a climber's body if he eats 0.70 kg of -15C snow which his body warms to body temperature of 37C.

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

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Part B

Calculate the energy absorbed from a climber's body if he melts 0.70 kg of -15C snow using a stove and drink the resulting 0.70 kg of water at 2C, which his body has to warm to 37C.

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/kgC, the latent heat of fusion is 333 kJ/kg, the specific heat of water is4186 J/kgC.

Part A

Calculate the energy absorbed from a climber's body if he eats 0.70 kg of -15C snow which his body warms to body temperature of 37C.

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

SubmitMy AnswersGive Up

Part B

Calculate the energy absorbed from a climber's body if he melts 0.70 kg of -15C snow using a stove and drink the resulting 0.70 kg of water at 2C, which his body has to warm to 37C.

Explanation / Answer

Specific heat capacity of snow (ice) { 2100 J/kg.C }

Specific heat capacity of water { 4186 J/kg.C }

Latent heat of fusion of ice { 333 kJ/kg }

Solve using:

Heat energy = mass * specific heat * change in temperature

Heat energy = mass * Latent heat of fusion

(a)

Total heat energy required =

Energy to raise temperature of 0.70kg of ice from -15°C to ice at 0 C + Energy to change 0.70 kg of ice at 0 C to 0.70 kg at water at 0 C + Energy to raise temperature of 0.70 kg of water from 0 C to 37 C

= (0.70 kg * 2100 J/kg.C * 15 C) + (0.70 kg * 333000 J/kg) + (0.70 kg * 4186 J/kg.C * 37C)

= 363567.4 J

(b)

Energy required to change 0.70 kg of water at 2 C to water at 37 C

= 0.70 kg * 4186 J/kg.C * (37 - 2)C

= 102557 J

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