It will take 1.11 min to heat the sample to its melting point.
Melting point = - 20°C
Boiling point = 85°C
∆H of fusion = 180 J/g
∆H of vap = 500 J/g
C(solid) = 1.0 J/g °C
C(liquid) = 2.5 J/g °C
C(gas) = 0.5 J/g °C
Mass of sample = 25 g
Initial temperature = - 40°C
Final temperature = 100°C
Rate of heating = 450 J/min
Specific heat capacity formula:- q = m ×C×∆T
Here, q = heat energy
m = mass
C = specific heat
∆T = temperature change
Melting point = - 20°C
C(solid) = 1.0 J/g °C
∆T = final temperature - initial temperature = -20 - (-40) = 20
Put these value in Specific heat capacity formula
q = m ×C×∆T
q = 25×1.0×20
=500J
The Rate of heating = 450 J/min
i.e. 450J = 1min
so, 500J = 1.11min
1.11 minutes does it take to heat the sample to its melting point.
The specific heat capacity is defined as the amount of heat absorbed in line with unit mass of the material whilst its temperature increases 1 °C.
Learn more about specific heat capacity here:- brainly.com/question/26866234
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There are three ways of abrasion occurs in nature; the wind, waves, and gravity. Abrasion is the mechanical process of wearing away of a rock surface and the movement of the particles while transporting through the wind, waves, and gravity.
Answer:
Explanation:
Temperature and thermal energy are in a direct proportion which means that if temperature of a substance increases, its thermal energy also increases and vice versa.
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<h3>~AH1807</h3>
Answer:
It was built to house the employees of the Chernobyl Nuclear Power Plants located 4 kilometers away and became the ninth nuclear city in the Soviet Union.
Explanation:
Answer:
The volume is increased.
Explanation:
According to <em>Charles' Law</em>, " <em>at constant pressure the volume and temperature of the gas are directly proportional to each other</em>". Mathematically this law is presented as;
V₁ / T₁ = V₂ / T₂ -----(1)
In statement the data given is,
T₁ = 10 °C = 283.15 K ∴ K = 273.15 + °C
T₂ = 20 °C = 293.15 K
So, it is clear that the temperature is being increased hence, we will find an increase in volume. Let us assume that the starting volume is 100 L, so,
V₁ = 100 L
V₂ = Unknown
Now, we will arrange equation 1 for V₂ as,
V₂ = V₁ × T₂ / T₁
Putting values,
V₂ = 100 L × 293.15 K / 283.15 K
V₂ = 103.52 L
Hence, it is proved that by increasing temperature from 10 °C to 20 °C resulted in the increase of Volume from 100 L to 103.52 L.
