Answer: 502 Joules
Explanation:
To calculate the mass of water, we use the equation:
![\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}](https://tex.z-dn.net/?f=%5Ctext%7BDensity%20of%20substance%7D%3D%5Cfrac%7B%5Ctext%7BMass%20of%20substance%7D%7D%7B%5Ctext%7BVolume%20of%20substance%7D%7D)
Density of water = 1 g/mL
Volume of water = 40.0 mL
Putting values in above equation, we get:
![1g/mL=\frac{\text{Mass of water}}{40.0mL}\\\\\text{Mass of water}=(1g/mL\times 40.0mL)=40.0g](https://tex.z-dn.net/?f=1g%2FmL%3D%5Cfrac%7B%5Ctext%7BMass%20of%20water%7D%7D%7B40.0mL%7D%5C%5C%5C%5C%5Ctext%7BMass%20of%20water%7D%3D%281g%2FmL%5Ctimes%2040.0mL%29%3D40.0g)
When metal is dipped in water, the amount of heat released by lead will be equal to the amount of heat absorbed by water.
![Heat_{\text{absorbed}}=Heat_{\text{released}}](https://tex.z-dn.net/?f=Heat_%7B%5Ctext%7Babsorbed%7D%7D%3DHeat_%7B%5Ctext%7Breleased%7D%7D)
The equation used to calculate heat released or absorbed follows:
![q=m\times c\times \Delta T](https://tex.z-dn.net/?f=q%3Dm%5Ctimes%20c%5Ctimes%20%5CDelta%20T)
q = heat absorbed by water
= mass of water = 40.0 g
= final temperature of water = 20.0°C
= initial temperature of water = 17.0°C
= specific heat of water= 4.186 J/g°C
Putting values in equation 1, we get:
![q=40.0\times 4.186\times (20.0-17.0)]](https://tex.z-dn.net/?f=q%3D40.0%5Ctimes%204.186%5Ctimes%20%2820.0-17.0%29%5D)
![q=502J](https://tex.z-dn.net/?f=q%3D502J)
Hence, the joules of heat were re-leased by the lead is 502