Problem 14.29 High-altitude mountain climbers do not eat snow, but always melt i
ID: 1510454 • Letter: P
Question
Problem 14.29
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 0.65 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.
3.38×105 J
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Part B
Calculate the energy absorbed from a climber's body if he melts 0.65 kg of -15C snow using a stove and drink the resulting0.65 kg of water at 2C, which his body has to warm to 37C.
Problem 14.29
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 0.65 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.
Q1 =3.38×105 J
SubmitMy AnswersGive Up
Answer Requested
Part B
Calculate the energy absorbed from a climber's body if he melts 0.65 kg of -15C snow using a stove and drink the resulting0.65 kg of water at 2C, which his body has to warm to 37C.
Explanation / Answer
a). Total heat energy required =
Energy to raise temperature of 0.65kg of ice from -15°C to ice at 0°C
+
Energy to change 0.65 kg of ice at 0°C to 0.65 kg at water at 0°C
+
Energy to raise temperature of 0.65kg of water from 0°C to 37°C
= (0.65 kg * 2.1 kJ/kg.°C * 15°C) + (0.65 kg * 333 kJ/kg) + (0.65 kg * 4.186 kJ/kg.°C * 37°C)
= 20.475 kJ + 216.45 kJ + 100.67 kJ
= 337.6 kJ
= 3.4 * 105 J
b). Energy required to change 1 kg of water at 2°C to water at 35°C
= 0.65 kg * 4.186 kJ/kg.°C * (37 - 2)°C
= 95.2 kJ
= 0.95 * 105 J