Answer:
The heat needed to warm 25.3 g of copper from 22°C to 39°C is 165.59 Joules.
Explanation:
Where:
Q = heat absorbed or heat lost
c = specific heat of substance
m = Mass of the substance
ΔT = change in temperature of the substance
We have mass of copper = m = 25.3 g
Specific heat of copper = c = 0.385 J/g°C
ΔT = 39°C - 22°C = 17°C
Heat absorbed by the copper :
The heat needed to warm 25.3 g of copper from 22°C to 39°C is 165.59 Joules.
A defining feature of a material that is a good insulator would be its ability to stop or cancel an electric charge of another material. This happens because the molecules in the material are stable and there are no free electrons that would allow the charges to flow.
Answer:
sulfur
Explanation:
sulfur because it have unknown metal
i just got done with the unit 2 weeks ago
Answer:
here:
Explanation:
The changes in temperature caused by a reaction, combined with the values of the specific heat and the mass of the reacting system, makes it possible to determine the heat of reaction.
Heat energy can be measured by observing how the temperature of a known mass of water (or other substance) changes when heat is added or removed. This is basically how most heats of reaction are determined. The reaction is carried out in some insulated container, where the heat absorbed or evolved by the reaction causes the temperature of the contents to change. This temperature change is measured and the amount of heat that caused the change is calculated by multiplying the temperature change by the heat capacity of the system.
The apparatus used to measure the temperature change for a reacting system is called a calorimeter (that is, a calorie meter). The science of using such a device and the data obtained with it is called calorimetry. The design of a calorimeter is not standard and different calorimeters are used for the amount of precision required. One very simple design used in many general chemistry labs is the styrofoam "coffee cup" calorimeter, which usually consists of two nested styrofoam cups.
When a reaction occurs at constant pressure inside a Styrofoam coffee-cup calorimeter, the enthalpy change involves heat, and little heat is lost to the lab (or gained from it). If the reaction evolves heat, for example, very nearly all of it stays inside the calorimeter, the amount of heat absorbed or evolved by the reaction is calculated.
(a) One form of the Clausius-Clapeyron equation is
ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂); where in this case:
Solving for ΔHv:
- ΔHv = R * ln(P₂/P₁) / (1/T₁ - 1/T₂)
- ΔHv = 8.31 J/molK * ln(5.3/1.3) / (1/358.96 - 1/392.46)
(b) <em>Normal boiling point means</em> that P = 1 atm = 101.325 kPa. We use the same formula, using the same values for P₁ and T₁, and replacing P₂ with atmosferic pressure, <u>solving for T₂</u>:
- ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂)
- 1/T₂ = 1/T₁ - [ ln(P₂/P₁) / (ΔHv/R) ]
- 1/T₂ = 1/358.96 K - [ ln(101.325/1.3) / (49111.12/8.31) ]
(c)<em> The enthalpy of vaporization</em> was calculated in part (a), and it does not vary depending on temperature, meaning <u>that at the boiling point the enthalpy of vaporization ΔHv is still 49111.12 J/molK</u>.
