The density of sample is 5 g/cm3
Given:
volume of sample = 20 cm3
mass of sample = 100 grams
To Find:
density of sample
Solution: Density is the measure of how much “stuff” is in a given amount of space. For example, a block of the heavier element lead (Pb) will be denser than the softer, lighter element gold (Au). A block of Styrofoam is less dense than a brick. It is defined as mass per unit volume
density = mass/volume
d = 100/20
d = 5 g/cm3
So, density of sample is 5 g/cm3
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Answer:
answer 2 because the baseball has less mass then the 1st one
Answer:
The amount of kilograms of ice at -20.0°C that must be dropped into the water to make the final temperature of the system 40.0°C = 0.0674 kg
Explanation:
Heat gained by ice in taking the total temperature to 40°C = Heat lost by the water
Total Heat gained by ice = Heat used by ice to move from -20°C to 0°C + Heat used to melt at 0°C + Heat used to reach 40°C from 0°C
To do this, we require the specific heat capacity of ice, latent heat of ice and the specific heat capacity of water. All will be obtained from literature.
Specific heat capacity of ice = Cᵢ = 2108 J/kg.°C
Latent heat of ice = L = 334000 J/kg
Specific heat capacity of water = C = 4186 J/kg.°C
Heat gained by ice in taking the total temperature to 40°C = mCᵢ ΔT + mL + mC ΔT = m(2108)(0 - (-20)) + m(334000) + m(4186)(40 - 0) = 42160m + 334000m + 167440m = 543600 m
Heat lost by water = mC ΔT = 0.25 (4186)(75 - 40) = 36627.5 J
543600 m = 36627.5
m = 0.0674 kg = 67.4 g of ice.
Answer:
Mass of the steel cube = 7800 kg
Volume of the steel = 1.025 cubic centimetre
Explanation:
Given:
The density of the steel = 7.8
Side of the cube = 12 cm
<u>(1)The mass of steel cube :</u>
We know that,
We are given with density and sides of the cube
then volume of the cube
= 
= 
= 1000 cubic centimetre
Now
mass = 7800 kg
<u>(2)volume of steel:</u>
Given the mass = 8 kg
Substituting the values
volume = 1.025 cubic centimetre
