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storchak [24]
3 years ago
13

if 980 KJ of energy are added to 6.2 L of water at 291 K, what will the final temperature of the water be

Chemistry
1 answer:
uranmaximum [27]3 years ago
6 0

Answer:

The final temperature of the water will  be 328.81 K .

Explanation:

Using the equation, <em>q = mcΔT</em>

here, <em>q = energy</em>

<em>m= mass</em>

<em>c= specific heat capacity </em>

<em>ΔT= change in temperature</em>

<em>Mass of water = 1kg (1000 g ) per liter</em>

<em>∴ 6.2 Liter of water = 6200 g</em>

<em>c of water ≈ 4.18 J /g/K</em>

<em>Now, </em>

<em>980000 = 6200*4.18*ΔT</em>

<em>ΔT = 37.81 K</em>

<em>∴ final temperature of the water = 291 + 37.81 = 328.81 K</em>

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