Answer: 502 Joules
Explanation:
To calculate the mass of water, we use the equation:
[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]
Density of water = 1 g/mL
Volume of water = 40.0 mL
Putting values in above equation, we get:
[tex]1g/mL=\frac{\text{Mass of water}}{40.0mL}\\\\\text{Mass of water}=(1g/mL\times 40.0mL)=40.0g[/tex]
When metal is dipped in water, the amount of heat released by lead will be equal to the amount of heat absorbed by water.
[tex]Heat_{\text{absorbed}}=Heat_{\text{released}}[/tex]
The equation used to calculate heat released or absorbed follows:
[tex]q=m\times c\times \Delta T[/tex]
q = heat absorbed by water
[tex]m[/tex] = mass of water = 40.0 g
[tex]T_{final}[/tex] = final temperature of water = 20.0°C
[tex]T_{initial[/tex] = initial temperature of water = 17.0°C
[tex]c[/tex] = specific heat of water= 4.186 J/g°C
Putting values in equation 1, we get:
[tex]q=40.0\times 4.186\times (20.0-17.0)][/tex]
[tex]q=502J[/tex]
Hence, the joules of heat were re-leased by the lead is 502
