# Assume you drank 240 mL of coffee at 80 degrees C and that this amount was cooled by your body to...

## Question:

Assume you drank 240 mL of coffee at 80 degrees C and that this amount was cooled by your body to the normal 37 degrees C. How many calories were supplied to your body? Assume that the coffee is essentially hot water with a density of 1.0 g/mL.

## Heat Transfer:

The heat exchange between two substances is related to the change in the temperature of both. We can quantify the heat lost or gained by one substance, q, to the change of its temperature, {eq}\displaystyle \Delta {/eq}T, using the equation, {eq}\displaystyle q = mc\Delta T {/eq}, where, m is its mass and c is its specific heat.

Determine the heat supplied to the body by finding the amount of heat lost by the coffee using the equation, {eq}\displaystyle q = -mc\Delta T {/eq}, where m is the mass of coffee, c = 1 cal/g{eq}\displaystyle ^\circ {/eq}C is the assumed specific heat of coffee, and the change in temperature is {eq}\displaystyle \Delta {/eq}T = 37 - 80 = -43{eq}\displaystyle ^\circ {/eq}C. We assign a negative sign since heat is lost by the coffee. We acquire the mass by simply multiplying the volume, V = 240 mL to the density, {eq}\displaystyle \rho {/eq} = 1.0 g/mL, or {eq}\displaystyle m = V\times \rho {/eq}. We proceed with the solution.

{eq}\begin{align} \displaystyle q &= -mc\Delta T\\ &= -V\times \rho \times c\times \Delta T\\ &= -240\ mL\times 1.0\ g/mL\times 1\ cal/g^\circ C\times -43 ^\circ C\\ &= \boxed{\rm 10320\ cal} \end{align} {/eq}

There is a total of {eq}\displaystyle \rm 10320\ cal {/eq} supplied to the body.