If you want to warm 3.10 g of water from 56.4 C to 82.4 C, how many calories would you need to add to the water?

The amount of calorie required to warm 3.10 g of water from 56.4°C to 82.4°C is 80.6 calories.

HOW TO CALCULATE CALORIES:

Q = m × c × ∆T

Where:

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80.6

Explanation:

q = mcAt

we have the grams of water, temp change, and specific heat of water is 4.186 J/g so we can plug in our numbers to find the heat energy in joules

q = (3.10)(4.184)(82.4-56.4)

q= 337.2304 J

we need our answer to be in calories so divide our answer by 4.184 and we get 80.6 calories

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