How Much Does Adding Ice Cool Your Coffee?

[Awer1]

1. An insulated Thermos contains 120 cm³ of hot coffee at 80.0°C. You put in a 19.0 g ice cube at its melting point to cool the coffee. By how many degrees (in Celsius) has your coffee cooled once the ice has melted and equilibrium is reached? Treat the coffee as though it were pure water and neglect energy exchanges with the environment. The specific heat of water is 4186 J/kg·K. The latent heat of fusion is 333 kJ/kg. The density of water is 1.00 g/cm³.



2. Tf = \frac{[(Ti)(Cw)(Mw) - (Hfus)(Mi)]}{Cw(Mw+Mi)}



3. Tf = (120cm3)(1.00x10^-3\frac{kg}{cm^3})(4186x10^-3\frac{kJ}{kg K})(80+273)K - (333\frac{kJ}{kg}x(19x10^-3kg))/(4186x10^-3\frac{kJ}{kg K})(19x10^-3kg)+(120x10^-3kg)

= 293.8743396K - 273K = 20.87433961^oC

The answer the book has.. : 21.809113626442^oC

After reworking the algebra the only way the book could of gotten that answer is if they didn't substitute any of the units and only changed the value 333 \frac{kJ}{kg} to a value of 333000. which gives an answer of 58.19088637^oC. which they then subtract it from the initial temp to get ..--> 80 - 58.19088637 = 21.80911363^oC. How is that even right? You have to convert everything so that the units match up right?.. in the books case how are the units being canceled out?

Here is my question: Who is right?

 

[Mapes]

The book is right. If you're adding 273 to all your temperatures to work in Kelvins, I don't see where you added 273 to the temperature of the ice. (Alternatively, you could just work in degrees Celsius.)