Question

Calculate the amount of heat water absorbs from a piece of hot metal using the following data: 75.0 g of cold water is placed in a calorimeter. The initial temperature of the water is 21.2 oC. To the calorimeter a 29.458 g piece of metal at 98.9 oC is added. The final temperature of the contents of the calorimeter is measured to be 29.5 oC. (HINT: the specific heat of water is 4.184 LaTeX: \frac{J}{g\cdot K}

Answer 1

Amount of heat absorbed by water is 2604.54 J.

Step-by-step explanation:

Amount of heat absorbed by water =
`m_(water)* C_(water)* \Delta T_(water)`

where m represents mass, C represents specific heat and
`\Delta T` represents change in temperature.

Here
`m_(water)=75.0` g ,
`C_(water)=4.184J/(g.^(0)\textrm{C})` and
`\Delta T` = (final temperature - initial temperature) = (29.5-21.2)
`^(0)\textrm{C}` = 8.3
`^(0)\textrm{C}`

So, amount of heat heat absorbed by water

=
`(75.0g)* (4.184\frac{J}{g.^(0)\textrm{C}})* (8.3^(0)\textrm{C})`

= 2604.54 J