Question

How to find the amount of heart energy transferred?

If a piece of aluminum metal with a mass of 65.5 hectograms has a temperature of 50.0 degrees Celsius and it is dropped into a swimming pool full of water at 25.0 degrees Celsius, how much heat energy was transferred to the pool water if the specific heat of aluminum is 0.8897 J/g * C ?

I know to use the equation q = mcΔT

but I'm a little lost with what to plug in as the initial and final temperature? Is it the temperature of the aluminium and the temperature of the swimming pool?

Answer 1

The amount of energy transferred to the pool water was 146 kJ.

Explanation:

We can assume that the mass of water in the pool is so large that the temperature of the water will not change significantly.

Then the final temperature of the aluminium will be the same as that of the pool (25.0 °C).

The formula for the quantity of heat `q` transferred is

`color(blue)(bar(ul(|color(white)(a/a)q = mCΔTcolor(white)(a/a)|)))" "`

where

`m color(white)(ll) =` the mass of the object
`C color(white)(m) =` its specific heat capacity
`ΔT =` its change in temperature

In this problem,

`m color(white)(ll)= 65.5 × 10^2color(white)(l) "g"`
`C color(white)(ll)= "0.8897 J·°C"^"-1""g"^"-1"`
`ΔT = T_text(f) - T_text(i) = "25.0 °C - 50.0 °C = -25.0 °C"`

`q = 65.5 × 10^2 color(red)(cancel(color(black)("g"))) × "0.8897 J"·color(red)(cancel(color(black)("°C"^"-1""g"^"-1"))) × "(-25.0)" color(red)(cancel(color(black)("°C"))) = "-1.46 × 10"^5 color(white)(l)"J" = "-146 kJ"`

The negative sign shows that heat was transferred from the aluminium to the pool water.