Determine the mass of iron heated to 85.0 °C to add to 54.0 grams of ice at 0.0 °C to produce water at 12.5 °C. The specific heat of iron is 0.045 J/g °C.

The correct answer is 6.39 kg.

Explanation:

Water is produced at 12.5 degree C, thus, the final temperature is 12.5 degree C.  

In the given case, q1 will be the heat required for the conversion of ice to water, and q2 will be the heat required to produced water exhibiting temperature 12.5 degree C.  

Thus, on the basis of the formula q1 and q2 are determined,  

q1 = ΔHf × m

= 334 J/g × 54 g = 18036 J

q2 = mCΔT

= 54 × 4.184 × 12.5 = 2824.2 J

Hence, the total heat required will be q1 + q2 = 18036 J + 2824.2 J

= 20860.2 K

The mass of iron can be calculated as:  

-qmetal = qwater

-m × c × ΔT = 20860.2

-m × 0.045 × (12.5 - 85) = 20860.2

m = 6393 g or 6.39 Kg

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idk what the question is about