Question

How many Joules of heat are required to raise the temperature of 20.0 grams of water from 30.0oC to 40.0oC?

Answer 1

The amount of heat required to raise the temperature of 20.0 grams of water from 30.0°C to 40.0°C is 8368 J.

Step-by-step explanation:

The specific heat of water is 4.184 J/g °C. To calculate the amount of heat required, you need to use the formula:

Q = m * c * ΔT

Where:
Q = heat transfer in Joules
m = mass of water in grams
c = specific heat capacity of water
ΔT = change in temperature

Using the given values, we have:
m = 20.0 g
c = 4.184 J/g °C
ΔT = (40.0 °C) - (30.0 °C) = 10.0 °C

Plugging in these values into the formula, we get:
Q = (20.0 g) * (4.184 J/g °C) * (10.0 °C) = 8368 J

Answer 2

840 J

Step-by-step explanation:

c ≈ 4200 J / (kg * °C)

m = 20 g = 0,02 kg

`t_(1)` = 30 °C

`t_(2)` = 40 °C

The formula is: Q = c * m * (
`t_(2) - t_(1)`)

Calculating:

Q = 4200 * 0,02 * (40 - 30) = 840 (J)