How much energy must be absorbed by 20.0 g of water to increase its temperature from 283.0 degrees c to 303.0 degrees c
1,672J
To calculate the amount of energy absorbed by the water, you can use the formula:
Q = mcΔT
where Q is the energy absorbed, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
The specific heat capacity of water is approximately 4.18 J/g·°C.
Let's calculate the energy absorbed by the water:
Q = (20.0 g) x (4.18 J/g·°C) x (303.0°C - 283.0°C)
Q = 20.0 g x 4.18 J/g·°C x 20.0°C
Q = 1672 J
Therefore, 20.0 g of water must absorb 1672 J of energy to increase its temperature from 283.0°C to 303.0°C.
To calculate the amount of energy needed to increase the temperature of a substance, we can use the specific heat capacity formula:
q = m * c * ΔT
Where:
q = heat energy gained or lost
m = mass of the substance
c = specific heat capacity
ΔT = change in temperature
In this case, the mass (m) of water is 20.0 g. The specific heat capacity (c) of water is approximately 4.18 J/g°C. And the change in temperature (ΔT) is from 283.0°C to 303.0°C, which is 20.0°C.
Substituting these values into the formula:
q = 20.0 g * 4.18 J/g°C * 20.0°C
Calculating:
q = 1672 J
Therefore, 1672 joules of energy must be absorbed by 20.0 g of water to increase its temperature from 283.0°C to 303.0°C.