When you drink cold water, your body must expend metabolic energy in order to maintain normal body temperature ( 37.0°C ) by warming up the water in your stomach. Could drinking ice water, then, substitute for exercise as a way to “burn calories?” Suppose you expend 430 kcalories during a brisk hour-long walk. How may litres of ice ( 0°C ) would you have to drink in order to use 430 kcalories of metabolic energy? For comparison, the stomach can hold about 1 litre .
The volume V of a mass m of water is given by V =m /ρ,where m is the mass, and ρ is the density of water. In order to warm a mass m of ice water to body temperature, the body must provide an amount Q of heat given by Q = cm∆T , where c is the specific heat of water and ΔT is 37 C°, the difference between body temperature (37 °C) and the temperature of ice water (0 °C) . Therefore using Q = cm∆T to calculate the required mass m of ice water, and V = m/ ρ to find the corresponding volume V.
The Attempt at a Solution
The volume of given water is V =m /ρ (1)
Rearranging Q = cm∆T in terms of m gives us: m= Q/c∆T (2)
Substiuting (2) into (1) yields: V = (Q/c∆T)/ρ = Qρ/c∆T (3)
The only problem with this is the units for this resulting equation (3) don't work out correctly and I am unsure what mathematical steps I am missing.
It's really the substitution that's confusing me.
The only way the units work out to be m^3 is if V = Q/ρc∆T where the density is part of the denominator which I cannot see how to get to.
I also realise there is another way to solve this problem (ie finding the heat energy needed for 1kg of ice water and dividing the 430kcal into this) but I'm just not sure what I'm missing in my method to reach the same answer.