Question

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)

Answer 1

Answer:
`2168 ^(\circ) \text{C}`

Step-by-step explanation:

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

From the question, we know that
`m=175, Q=1.57 * 10^(6)`. Furthermore, we know that
`c=4.18` (this is just a fact).

So, we get that
`1.57 * 10^(6)=175(4.18)(\triangle T)`, meaning
`\triangle T=2146`.

Thus,
`t_(f)-22.0=2146 \longrightarrow t_(f)=\boxed{2168 ^(\circ) \text{C}}`