Question

Calculate energy absorbed from your body if you (a) eat 1.0 kg of −10 oC snow which your body warms to a body temperature of 37 oC. (b) You melt 1.0 kg of −10 oC snow using a stove and drink the resulting 1.0 kg of water at 2 oC, which your body has to warm to 37 oC.

Answer 1

The amount of energy absorbed from your body if you eat 1.0 kg of -10°C snow and if you melt 1.0 kg of -10°C snow and drink the resulting water can be calculated. For eating the snow, the total energy absorbed would be 510.96 kJ. For melting the snow, the total energy absorbed would be 530.86 kJ.

Step-by-step explanation:

The amount of energy absorbed from your body can be calculated using the specific heat of snow, the heat of fusion of ice, and the specific heat of water. Let's calculate for each situation:

a) Eating snow:

1. First, we need to calculate the heat required to warm the snow from -10°C to 0°C using the specific heat of snow. Q = mass x specific heat x change in temperature. Q = 1.0 kg x 2.09 kJ/kg°C x 10°C = 20.9 kJ
2. Next, we need to calculate the heat required to melt the ice at 0°C using the heat of fusion of ice. Q = mass x heat of fusion. Q = 1.0 kg x 334 kJ/kg = 334 kJ
3. Finally, we need to calculate the heat required to warm the water from 0°C to 37°C using the specific heat of water. Q = mass x specific heat x change in temperature. Q = 1.0 kg x 4.18 kJ/kg°C x 37°C = 155.06 kJ

The total energy absorbed is 20.9 kJ + 334 kJ + 155.06 kJ = 510.96 kJ. Therefore, by eating 1.0 kg of -10°C snow, your body would absorb 510.96 kJ of energy.

b) Melting snow:

1. First, we need to calculate the heat required to warm the water from -10°C to 0°C using the specific heat of water. Q = mass x specific heat x change in temperature. Q = 1.0 kg x 4.18 kJ/kg°C x 10°C = 41.8 kJ
2. Next, we need to calculate the heat required to melt the ice at 0°C using the heat of fusion of ice. Q = mass x heat of fusion. Q = 1.0 kg x 334 kJ/kg = 334 kJ
3. Finally, we need to calculate the heat required to warm the water from 0°C to 37°C using the specific heat of water. Q = mass x specific heat x change in temperature. Q = 1.0 kg x 4.18 kJ/kg°C x 37°C = 155.06 kJ

The total energy absorbed is 41.8 kJ + 334 kJ + 155.06 kJ = 530.86 kJ. Therefore, by melting 1.0 kg of -10°C snow and drinking the resulting water, your body would absorb 530.86 kJ of energy.

Answer 2

(a) 164.5 kJ.

(b) 122.5 kJ

Step-by-step explanation:

Specific Heat Capacity: The specific heat capacity of a substance, is the quantity of heat required to raise the temperature of a unit mass (1 kg) of a substance through a degree rise in temperature (1 K). It is expressed in J/kgK

Mathematically, it can be expressed as

Q = cmΔT.

Where Q = quantity of heat, c = specific heat capacity, m = mass of the substance, ΔT = change in temperature

(a) Q = cmΔT

Where c = specific heat capacity of the body = 3.5 kJ/kgK = 3500 J/kgK, m = 1.0 kg,

ΔT = T₂ - T₁ = 37-(-10) = 37+10 = 47 °C

Substituting this values in the equation above,

Q = 3500×1.0×47

Q = 164500 J

Q = 164.5 kJ.

Therefore, Energy absorbed by the body = 164.5 kJ

(b) Q =cmΔT

Where Q = quantity of heat, c = specific heat capacity of the the body = 3.5 kJ/kgK = 3500 J/kgK, m = 1.0 kg,

ΔT = T₂ - T₁ = 37 -2 = 35 °C.

∴ Q = 3500×1.0×35

Q = 122500 J

Q = 122.5 kJ

∴ Energy absorbed by the body = 122.5 kJ