Wu Ming  wrote:   
   > Energy payback: W1*(t-tcool)+Wh =W2*t   
   >   
   > Where:   
   > W1 is the average wattage at lower T   
   > W2 is the average wattage at 98°   
   > t is the number of hours   
   > tcool is the time to cool down from 98° to T   
   > Wh is the energy required to re-heat from T to 98°   
   >   
   > W1, W2 are known from my previous   
   > tcool is unknown and I tried to guesstimate it   
   > Wh is calculated as:   
   >   
   > T° C to 98° C	Wh	Wh *1.25   
   > 90	37.24	46.55   
   > 80	83.8	104.75   
   > 60	176.91	221.14   
   > After 6 hrs 21°	358.48	448.1   
   >   
   > Second column from an online calculator. Wolfram Alpha as example. Third   
   > accounts for the not perfect energy transfer.   
   >   
   > Solving for t is simple. Then the fun begins.   
   >   
      
   Formatting appears to be a mess. On my client at least. Doesn’t know how to   
   keep LF.   
      
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