Question

1. Heat opens capillaries and improves blood flow.  The reverse is true too:  cold capillaries close.   Thus, for a black eye where you want to prevent blood buildup causing painful swelling, you use ice. 

Now consider a patient who is told to keep hot compresses on an eye infection for 10 minutes.  She discovers that her compress is no longer hot after only 5 minutes and therefore wants to keep it warm twice as long.   Is the better strategy to use more hot water to keep it warm longer, or use the same amount of water as before, but just make the water hotter?

2. The desert sand is very hot during the day and very cold at night. What does this tell you about its specific heat capacity?

3. James Joule used a spinning set of paddles to heat the water in which they were placed, and by comparing the mechanical energy he put in, and temperature rise of the liquid afterwards, he determined the interconversion between mechanical and thermal energy.    Inspired by Joule's experiment, you decide to heat your bath water by pushing your hand through it in circles. Estimate the total distance your hand will have travelled to raise the water temperature by 10°C in a typical bathtub. You may assume your hand exerts a continuous force of 50 N.

4. When cooking frozen cheese ravioli, the directions say to put the 255 grams of cheese-filled pasta into 3 quarts of boiling water. We want to explore, by calculations, why you are told to use 3 quarts when it's obvious that 1 quart would easily cover them all, and get dinner cooked even faster? 

Suppose the ravioli are in the freezer at - 40 °C. You may consider the ravioli to have a specific heat of 0.4 cal/gam-°C, both when frozen and when in water.

(A) By how much does the temperature of the 3 quarts of water drop when you add the frozen ravioli?

(B)  How much would the water temperature drop if you used only 1 quart?

(C) So what is the answer?  Why do they ask for 3 quarts?

5. The energy of a thunderstorm results from the condensation of water vapor in humid air. Suppose a thunderstorm could condense all the water vapor in 10 km³ of air.

How much heat does this release? 

(You may assume each cubic meter of air contains 0.017 kg of water vapor.)

How does this compare to an atomic bomb which releases an energy of 2 x 10¹⁰ kcal?

Answer 1

Newton's law of cooling

\(\frac{d Q}{d h}=h A\left(T-T_{\text {env }}\right)\) where \(d Q=-m e d T\)

Solution if \(T_{f}=\)  \({T_{\text {env }}} + \left(T_{0}\right. - {T_{\text {env }}}\)) \(e^{-K t}\) ,where \(k=\frac{h A}{m c}\)

\(h\) is the heat transfer coefficient

A is the heat transfer sorface area mis the mass and "C" the specific heat

clear time "t"from solution: When T \(=0\)

\(\frac{-T_{\text {env }}}{T_{0}-T_{\text {env }}}-\frac{T_{\text {env }}}{\text { Tenv } - T_{0}}=e^{-k t} \Rightarrow-k t=\ln \left(\frac{\text { Tenv }}{\text { Tenv }-T_{0}}\right)\)

\(t=\frac{1}{k} \ln \left(\frac{{T_{\text {env }}}-T_{0}}{T_{\text {env }}}\right) \Rightarrow t=\frac{m c}{h A} \ln \left(\frac{{T_{\text {env }}}-T_{0}}{{T_{\text {env }}}}\right)\)

To get \(2 t\) we need \(2 m\) not \(2 t_{0}\). So the best way is increase the mass of water