sum of heats gained is zero
13oz*28.3g/oz*4.18J/gC*deltatemp+33oJ/g*massice=0
now for delta temp, the change in F is 71-39=22F, and in C that is 22*5/9
solve for mass of ice.
When ice melts, it absorbs 0.33 kJ per gram.
How much ice is required to cool a 13.0 oz drink from 71 ∘F to 39 ∘F, if the heat capacity of the drink is 4.18 J/g∘C? (Assume that the heat transfer is 100 % efficient.)
2 answers
I actually found the answer. The answer is 83 grams.
You must convert the values accordingly. oz to grams, Fahrenheit to Celsius.
You calculate the heat lost by the drink first. This is the setup:
=(368.544g)(4.18J/gC)(-17.7778C)
the change in temp is found by first converting your values to Celsius, then subtracting the initial from the final.
after you find the answer to the above equation, you divide it by the 0.33kJ that ice absorbs when it melts. You must convert the kJ to J first though.
this will give you the number of grams needed to cool the drink.
You must convert the values accordingly. oz to grams, Fahrenheit to Celsius.
You calculate the heat lost by the drink first. This is the setup:
=(368.544g)(4.18J/gC)(-17.7778C)
the change in temp is found by first converting your values to Celsius, then subtracting the initial from the final.
after you find the answer to the above equation, you divide it by the 0.33kJ that ice absorbs when it melts. You must convert the kJ to J first though.
this will give you the number of grams needed to cool the drink.
9 answers