How Much Ice to Cool Boiling Tea to 65C?

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To cool 200 grams of boiling tea to 65°C, approximately 45.9 grams of ice at -15°C is required. The calculation involves the ice absorbing 7000 calories of heat, factoring in its specific heat capacity of 0.5 cal/g°C and the latent heat of fusion at 80 cal/g. The equation combines the heat absorbed by the ice as it warms to 0°C, melts, and then the resulting water warms to 65°C. The solution appears accurate based on the calculations provided. This method effectively demonstrates the principles of heat transfer and specific heat capacity.
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Homework Statement


your 200 gram cup of tea is boiling hot. About how much ice should you add to bring it down to 65C? Ice is initially at -15C. The specific heat capacity of ice id .5cal/gC

The Attempt at a Solution


So the ice we add needs to absorb 7000 calories of heat.
Heat capacity is Q/T heat divided by change in temperature . And specific heat capacity is is heat capacity divided by mass.
So to heat the ice to 0 Celsius we have
15C(.5cal)/gC(x) + 65(x) + 80 cal/g(x) =7000 calories.
cleaning it up
(15)(.5)x+65x+80x=7000 in units of energy
the 80 comes from the latent heat for melting ice which is 80cal/gram
and x is the unknown mass of ice.
and i get that x=45.9 grams of ice.
 
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It looks correct.

ehild
 

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