If 8.00 kJ of heat are added to 200. g of water
at 20.0 ◦C, what is the final temperature of
the water
Hydrogen peroxide decomposes into water
and oxygen when exposed to heat or light.
A tightly capped bottle of hydrogen peroxide
is placed on a balance and exposed to light for
three weeks. The mass reading on the balance
does not change. This is an example of
You should make your posts separate. They have a better chance of being answered that way.
#1,
q = mass x specific heat x (Tfinal-Tinitial)
#2.
Conservation of mass.
To find the final temperature of the water after adding 8.00 kJ of heat, we can use the equation:
q = m * c * ΔT
Where:
q is the heat energy added,
m is the mass of the water,
c is the specific heat capacity of water, and
ΔT is the change in temperature.
We first need to calculate the heat capacity (C) of the water:
C = m * c
Given:
m = 200. g
c = 4.18 J/g°C (specific heat capacity of water)
C = (200. g) * (4.18 J/g°C)
C = 836 J/°C
Now we can calculate the change in temperature (ΔT):
q = C * ΔT
Given:
q = 8.00 kJ = 8.00 * 1000 J (since 1 kJ = 1000 J)
8.00 * 1000 J = 836 J/°C * ΔT
ΔT = (8.00 * 1000 J) / (836 J/°C)
ΔT = 9.57 °C
Finally, we can find the final temperature by adding the change in temperature to the initial temperature:
Final temperature = Initial temperature + ΔT
Final temperature = 20.0 °C + 9.57 °C
Final temperature = 29.57 °C
Therefore, the final temperature of the water after adding 8.00 kJ of heat is approximately 29.57 °C.
To determine the final temperature of the water, we can use the equation:
q = m * c * ΔT
where:
q is the amount of heat transferred,
m is the mass of the substance,
c is the specific heat capacity of the substance,
ΔT is the change in temperature.
In this case, the heat transferred (q) is given as 8.00 kJ, the mass (m) is 200 g, and the initial temperature (T1) is 20.0 °C. We need to find the final temperature (T2).
First, let's convert the given heat from kilojoules (kJ) to joules (J):
8.00 kJ * 1000 J/1 kJ = 8000 J
Next, we need to find the specific heat capacity of water (c). The specific heat capacity of water is usually around 4.18 J/g°C.
Now we can rearrange the equation to solve for ΔT:
ΔT = q / (m * c)
ΔT = 8000 J / (200 g * 4.18 J/g°C)
ΔT ≈ 9.57 °C
Finally, we can find the final temperature (T2) by adding the change in temperature (ΔT) to the initial temperature (T1):
T2 = T1 + ΔT
T2 = 20.0 °C + 9.57 °C
T2 ≈ 29.6 °C
Therefore, the final temperature of the water is approximately 29.6 °C.