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

Answer:

Amount of heat absorbed by water is 2604.54 J.

Explanation:

Amount of heat absorbed by water = $m_{water}\times C_{water}\times \Delta T_{water}$

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

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

     = 2604.54 J