Question

1. How much energy is required to heat 32.5g of water from 34°C to 75°C?

Answer 1

`\boxed {\boxed {\sf 5569.85 \ Joules}}`

Step-by-step explanation:

We are asked to find how much energy is required to heat a sample of water.

We will use the following equation:

`q= mc \Delta T`

The mass of the water is 32.5 grams. The specific heat capacity of water is 4.18 Joules per gram degree Celsius.

The change in temperature is the difference between the final temperature and the initial temperature. The water's temperature is raised from 34 degrees Celsius to 75 degrees Celsius.

• ΔT= final temp - inital temp
• ΔT = 75 °C - 34 °C = 41 °C

Now we know all three variables:

• m= 32.5 g
• c= 4.18 J/g °C
• ΔT = 41 °C

Substitute the values into the formula.

`q= (32.5 \ g)(4.18 \ J/g \textdegree C)(41 \ \textdegree C)`

Multiply the first 2 numbers together. The units of grams cancel.

`q= 135.85 \ \ J/ \textdegree C(41 \ \textdegree C)`

Multiply again. This time, the units of degrees Celsius cancel.

`q= 5569.85 \ J`

5,569.85 Joules of energy are required.

Answer 2

5.850 kJ (of heat energy)

Step-by-step explanation:

What we need:

[] The mass of the material = m = 32.5g

[] The temperature change that occurs = ΔT = 43

() 75 - 34 = 43

[] The specific heat capacity of the material = c

() This is the amount of heat required to raise 1 gram of that substance by 1°C

() For water, this is 4.186Jg°C

The equation to use:

Q = m * c * ΔT

Q = 32.5 * 4.186 * 43

Q = 5849.935 J

Q = 5.850 kJ

Have a nice day!

I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)

- Heather