Question

Water (3070 g ) is heated until it just begins to boil. if the water absorbs 5.03×105 j of heat in the process, what was the initial temperature of the water?

Answer 1

Answer:
`60.8\°C`

Explanation:

The heat absorbed
`Q` is given by:

`Q=(m) (c) \Delta T` (1)

Where:

`Q=5.03 (10)^(5) J`

`m=3070 g` is the mass of water

`c=4.18 (J)/(g \°C)` is the specific heat of water

`\Delta T=T_(f)-T_(i)` is the variation in temperature, being the final temperature
`T_(f)=100\°C` which is the boiling temperature of water

`T_(i)` is the initial temperature

Finding
`\Delta T`:

`\Delta T=(Q)/((m)(c))` (2)

`\Delta T=(5.03 (10)^(5) J)/((3070 g)(4.18 (J)/(g \°C)))` (3)

`\Delta T= 39.19 \°C` (4)

Knowing
`\Delta T` and
`T_(f)` we can find
`T_(i)`:

`\Delta T=T_(f)-T_(i)`

`T_(i)=T_(f) - \Delta T` (5)

`T_(i)=100 \°C - 39.19 \°C` (6)

Finally:

`T_(i)=60.8 \°C` This is the initial temperature of the water