Answer:
There are three main sources of heat in the deep earth: (1) heat from when the planet formed and accreted, which has not yet been lost; (2) frictional heating, caused by denser core material sinking to the center of the planet; and (3) heat from the decay of radioactive elements.
Answer:
A = -213.09°C
B = 15014.85 °C
C = -268.37°C
Explanation:
Given data:
Initial volume of gas = 5.00 L
Initial temperature = 0°C (273 K)
Final volume = 1100 mL, 280 L, 87.5 mL
Final temperature = ?
Solution:
Formula:
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Conversion of mL into L.
Final volume = 1100 mL/1000 = 1.1 L
Final volume = 87.5 mL/1000 = 0.0875 L
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
T₂ = V₂T₁ / V₁
T₂ = 1.1 L × 273 K / 5.00 L
T₂ = 300.3 L.K / 5.00 K
T₂ = 60.06 K
60.06 K - 273 = -213.09°C
2)
V₁/T₁ = V₂/T₂
T₂ = V₂T₁ / V₁
T₂ = 280 L × 273 K / 5.00 L
T₂ = 76440 L.K / 5.00 K
T₂ = 15288 K
15288 K - 273 = 15014.85 °C
3)
V₁/T₁ = V₂/T₂
T₂ = V₂T₁ / V₁
T₂ = 0.0875 L × 273 K / 5.00 L
T₂ = 23.8875 L.K / 5.00 K
T₂ = 4.78 K
4.78 K - 273 = -268.37°C
The true statement is (A) energy is absorbed during the reaction
For the products to have more energy, they must absorb it from the surrounding.
Answer : The specific heat of the substance is 0.0936 J/g °C
Explanation :
The amount of heat Q can be calculated using following formula.

Where Q is the amount of heat required = 300 J
m is the mass of the substance = 267 g
ΔT is the change in temperature = 12°C
C is the specific heat of the substance.
We want to solve for C, so the equation for Q is modified as follows.

Let us plug in the values in above equation.


C = 0.0936 J/g °C
The specific heat of the substance is 0.0936 J/g°C
The question is asking to choose among the following choices is cannot be considered as a single phase and base on my further research and understanding about the sad topic, I would say that the answer would be <span>d) a heterogeneous mixture. I hope you are satisfied with my answer </span>