We can rephrase the statement with a little more specificity in order to understand the answer here.
The mass of the products can never be more than the The mass that is expected.
Answer:
-10778.95 J heat must be removed in order to form the ice at 15 °C.
Explanation:
Given data:
mass of steam = 25 g
Initial temperature = 118 °C
Final temperature = 15 °C
Heat released = ?
Solution:
Formula:
q = m . c . ΔT
we know that specific heat of water is 4.186 J/g.°C
ΔT = final temperature - initial temperature
ΔT = 15 °C - 118 °C
ΔT = -103 °C
now we will put the values in formula
q = m . c . ΔT
q = 25 g × 4.186 J/g.°C × -103 °C
q = -10778.95 J
so, -10778.95 J heat must be removed in order to form the ice at 15 °C.
Sedimentary rock can be minazute by adding more heat n pressure
Answer:
1. ΔE = 0 J
2. ΔH = 0 J
3. q = 3.2 × 10³ J
4. w = -3.2 × 10³ J
Explanation:
The change in the internal energy (ΔE) and the change in the enthalpy (ΔH) are functions of the temperature. If the temperature is constant, ΔE = 0 and ΔH = 0.
The gas initially occupies a volume V₁ = 20.0 L at P₁ = 3.2 atm. When the pressure changes to P₂ = 1.6 atm, we can find the volume V₂ using Boyle's law.
P₁ × V₁ = P₂ × V₂
3.2 atm × 20.0 L = 1.6 atm × V₂
V₂ = 40 L
The work (w) can be calculated using the following expression.
w = - P . ΔV
where,
P is the external pressure for which the process happened
ΔV is the change in the volume
w = -1.6 atm × (40L - 20.0L) = -32 atm.L × (101.325 J/1atm.L) = -3.2 × 10³ J
The change in the internal energy is:
ΔE = q + w
0 = q + w
q = - w = 3.2 × 10³ J
Guess and check, test, trial and error, completion.
