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 is4186 J/kg?C?.

PART A) Calculate the energy absorbed from a climber's body if he eats 0.40kg of -15?C snow which his body warms to body temperature of 37?C.

PART B) Calculate the energy absorbed from a climber's body if he melts 0.40kg of -15?C snow using a stove and drink the resulting 0.40kg of water at 2?C, which his body has to warm to 37?C.

Answer 1

A)

Total Energy absorbed by body = Energy required to heat ice to 0 C + Energy required to melt ice + Energy required to heat water to 37 C

= m c dT + m L + mc dT

= 0.40 (2100 * (0-(-15)) + 333* 10^3 + 4186 * (37-0) )

= 0.40 * 519382 = 207752.8 J = 207.7528 kJ

B)

Total Energy absorbed from body = Energy required to heat water to 37 C by body

= 0.40 * 4186 * (37-2) = 58604 J = 58.604 kJ

Answer 2

1. Heat absorbed from the body= m[ C_icedT_ice +Heat of fusion +C_wterdT_water]

=0.40*[2100*(0-(-15)+333+4186*(37-0)]

=0.40*186715=74686 J=74.686 kJ

1. Now heat absorbed= mC_waterdT_water

=0.4*4186*(37-2)

=20804 J=20.804 kJ

Answer 3

A) Energy absorbed = 0.4kg (2100*15 J + 333 kJ + 4186*37) = 207.7528 kJ

B) Energy absorbed = 0.4kg (4186*(37 - 2)) = 58.604 kJ

Answer 4

q=c*m*(t1-t2)
specific heat of water is about: 4,186 joule/kilogram