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April 19, 2014

April 19, 2014

Posted by **James** on Friday, February 17, 2012 at 1:58pm.

"The idea comes to mind to put some hot rocks in your frozen bath water. What mass of rocks at 220 degrees Celsius needs to be added to 500.0L of frozen bath water at 0 degrees celsius to raise its temperature to 55 degrees celsius? The specific heat of the rcks is 2.5cal/g-C* and that of water is 1cal/g-C*. The heat of fusion for ice is 80cal/g. Assume there is no heat transfer between the ice/water and the walls of your bath.

Any help is GREATLY appreciated

- Chemistry -
**DrBob222**, Friday, February 17, 2012 at 3:40pmIt appears that ALL of the bath water is frozen and not just a part of it.

heat lost by rocks + heat absorbed by

ice melting + heat absorbed by liquid water to raise T to final T = 0

heat lost by rocks is

mass rocks x specific heat rocks x (Tfinal-Tinitial)

heat absorbed by ice to melt it.

mass ice x delta Hfusion

heat to raise temperature of liquid water.

mass water x specific heat H2O x (Tfinal-Tinitial)

Just add all of these in a string and solve for mass rocks. In a string it looks like this

[mass rocks x sp.h. rocks x (Tf-Ti)] + (mass ice x delta Hfusion) + [mass melted ice x sp.h H2O x (Tf-Ti)] = 0

Substitute and solve for mass rocks. The final answer is a large number.

- Chemistry -
**James**, Friday, February 17, 2012 at 5:04pmTHANK YOU!

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