I am confused about how to do this one. Thanks! An insulated beaker with negligi

Question

I am confused about how to do this one. Thanks!

An insulated beaker with negligible mass contains liquid water witha mass of 0.340 kg and a temperature of 79.0 degrees Celsius.

How much ice at a temperature of-22.0 degrees Celsius must be dropped into the water so that thefinal temperature of the system will be 39.0 Celsius?

[Take the specific heat of liquidwater to be 4190 K, the specific heat of ice to be 2100 J/kg*K andthe heat of fusion for water to be 334 kJ/kg.]
How much ice at a temperature of-22.0 degrees Celsius must be dropped into the water so that thefinal temperature of the system will be 39.0 Celsius?

[Take the specific heat of liquidwater to be 4190 K, the specific heat of ice to be 2100 J/kg*K andthe heat of fusion for water to be 334 kJ/kg.]

Explanation / Answer

What you need to find out first is how much energy will berequired to lower the temperature that much. For the water todrop to 39degrees you constants are written wrong. ( specific heatof liquid water to be 4.190 J/g*C, the specific heat of ice to be2.100 J/g*C and the heat of fusion for water to be 334 J/g) 4.186 J/g*C *340 g * (79-39C)=56929.6 J in total needed now for the Ice you must find how much energy is absorbed pergram of ice. 22C*2.1J/gC*1g+39C*4.186J/gC*1g+334J=543.454J/g ofice. divide the top number by the bottom to get the number of gramsneeded. 56929.6J/543.454J/g=104.8 g of ice

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