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4. When 175 g of water is heated from 22.0°C, 1.57x106 J of energy are produced.
What is the final temperature of the water? (AT=Tr-Ti)

Sagot :

Medunno13

Answer: [tex]2168 ^{\circ} \text{C}[/tex]

Explanation:

For this question, we can use the formula [tex]Q=mc \triangle T[/tex], where [tex]Q[/tex] is the amount of heat absorbed, [tex]m[/tex] is the mass of the sample, [tex]c[/tex] is the specific heat constant, and [tex]\triangle T[/tex] is the change in temperature (final temperature minus initial temperature as stated in the question).

From the question, we know that [tex]m=175, Q=1.57 \times 10^{6}[/tex]. Furthermore, we know that [tex]c=4.18[/tex] (this is just a fact).

So, we get that [tex]1.57 \times 10^{6}=175(4.18)(\triangle T)[/tex], meaning [tex]\triangle T=2146[/tex].

Thus, [tex]t_{f}-22.0=2146 \longrightarrow t_{f}=\boxed{2168 ^{\circ} \text{C}}[/tex]

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