Posted by **mar13** on Sunday, December 4, 2011 at 7:58am.

An insulated Thermos contains 140 cm3 of hot coffee at 89.0°C. You put in a 11.0 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.

- physics -
**drwls**, Sunday, December 4, 2011 at 9:00am
At equilbrium, the melted ice (mass m) and the water it changes into will gain as much heat as the original coffee (mass M) loses.

0.140g*(89 - T)*4186

= 0.011*[333,000 + T*4186]

52,160 -586 T = 3663 + 46 T

632 T = 48,497

T = 77 C

Solve for final temperature T

- physics -
**Shahiahi**, Saturday, June 23, 2012 at 10:54pm
Heat releasing from coffee (water in this case) is equal to the heat gain for ice melting to water at 0 deg plus water to raise for the final temp.

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