Answer: 4Kcal
Explanation:
H= mcø
M=200g
C= 1 cal/g/°c
Ø= 40-20=20°c
H= 200*1*20= 4000calories= 4Kcal
Answer:
384.2 K
Explanation:
First we convert 27 °C to K:
- 27 °C + 273.16 = 300.16 K
With the absolute temperature we can use <em>Charles' law </em>to solve this problem. This law states that at constant pressure:
Where in this case:
We input the data:
300.16 K * 1600 m³ = T₂ * 1250 m³
And solve for T₂:
T₂ = 384.2 K
Answer:
15.69 dozen
Explanation:
Mass of penny = 5 g
Dozens of penny =..?
Next, we shall convert 5 g to gross. This can be obtained as follow:
3824 g = 1000 gross
Therefore,
5 g = 5 g × 1000 gross / 3824 g
5 g = 1.3075 gross
Thus, 5 g is equivalent to 1.3075 gross.
Finally, we convert 1.3075 gross to dozen. This can be obtained as follow:
1 gross = 12 dozen
Therefore,
1.3075 gross = 1.3075 gross × 12 dozen / 1 gross
1.3075 gross = 15.69 dozen
Thus, 5 g of penny is equivalent to 15.69 dozen
Answer:
Explanation:
Combustion reaction is given below,
C₂H₅OH(l) + 3O₂(g) ⇒ 2CO₂(g) + 3H₂O(g)
Provided that such a combustion has a normal enthalpy,
ΔH°rxn = -1270 kJ/mol
That would be 1 mol reacting to release of ethanol,
⇒ -1270 kJ of heat
Now,
0.383 Ethanol mol responds to release or unlock,
(c) Determine the final temperature of the air in the room after the combustion.
Given that :
specific heat c = 1.005 J/(g. °C)
m = 5.56 ×10⁴ g
Using the relation:
q = mcΔT
- 486.34 = 5.56 ×10⁴ × 1.005 × ΔT
ΔT= (486.34 × 1000 )/5.56×10⁴ × 1.005
ΔT= 836.88 °C
ΔT= T₂ - T₁
T₂ = ΔT + T₁
T₂ = 836.88 °C + 21.7°C
T₂ = 858.58 °C
Therefore, the final temperature of the air in the room after combustion is 858.58 °C