How much heat is required to convert 74 grams of ice at -4 C into 74 grams of water at 52 C?

First you have to warm the ice from -4 c to 0 c

 

74 gm * .5 cal/gm-c  * (4 c)  = 148 cal =   619.23 j         (1 cal = 4.184 j )

 

 

Then you have to melt the ice....this occurs at a constant temperature of 0 c

 

74 gm * 334 J/g  =   24716 j

 

Then you have to warm the water (which is now at 0 c) to 52 c

 

74 *1 * 52 = 3848 calories =16100 j

 

619.23 + 24716 + 16100 j = 41435.3 j =  41.4353 kj