Question

A lead mass is heated and placed in a foam cup calorimeter containing 40.0 mL of water at 17.0°C. The water reaches a temperature of 20.0°C. How many joules of heat were re-leased by the lead?

Answer 1

Answer: 502 Joules

Step-by-step explanation:

To calculate the mass of water, we use the equation:

`\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}`

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* 40.0mL)=40.0g`

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}}`

The equation used to calculate heat released or absorbed follows:

`q=m* c* \Delta T`

q = heat absorbed by water

`m` = mass of water = 40.0 g

`T_(final)` = final temperature of water = 20.0°C

`T_{initial` = initial temperature of water = 17.0°C

`c` = specific heat of water= 4.186 J/g°C

Putting values in equation 1, we get:

`q=40.0* 4.186* (20.0-17.0)]`

`q=502J`

Hence, the joules of heat were re-leased by the lead is 502