Answer:
The specific heat of the metal is 2.09899 J/g℃.
Explanation:
Given,
For Metal sample,
mass = 13 grams
T = 73°C
For Water sample,
mass = 60 grams
T = 22°C.
When the metal sample and water sample are mixed,
The addition of metal increases the temperature of the water, as the metal is at higher temperature, and the addition of water decreases the temperature of metal. Therefore, heat lost by metal is equal to the heat gained by water.
Since, heat lost by metal is equal to the heat gained by water,
Qlost = Qgain
However,
Q = (mass) (ΔT) (Cp)
(mass) (ΔT) (Cp) = (mass) (ΔT) (Cp)
After mixing both samples, their temperature changes to 27°C.
It implies that
, water sample temperature changed from 22°C to 27°C and metal sample temperature changed from 73°C to 27°C.
Since, Specific heat of water = 4.184 J/g°C
Let Cp be the specific heat of the metal.
Substituting values,
(13)(73°C - 27°C)(Cp) = (60)(27°C - 22℃)(4.184)
By solving, we get Cp =
Therefore, specific heat of the metal sample is 2.09899 J/g℃.
<span>S-shadows tells that the earths interior is liquid.</span>
Answer:
The amount of work done on the system is 18234 J and the final positive sign means that this work corresponds to an increase in internal energy of the gas.
Explanation:
Thermodynamic work is called the transfer of energy between the system and the environment by methods that do not depend on the difference in temperatures between the two. When a system is compressed or expanded, a thermodynamic work is produced which is called pressure-volume work (p - v).
The pressure-volume work done by a system that compresses or expands at constant pressure is given by the expression:
W system= -p*∆V
Where:
- W system: Work exchanged by the system with the environment. Its unit of measure in the International System is the joule (J)
- p: Pressure. Its unit of measurement in the International System is the pascal (Pa)
- ∆V: Volume variation (∆V = Vf - Vi). Its unit of measurement in the International System is cubic meter (m³)
In this case:
- p= 10 atm= 1.013*10⁶ Pa (being 1 atm= 101325 Pa)
- ΔV= 2 L- 20 L= -18 L= -0.018 m³ (being 1 L=0.001 m³)
Replacing:
W system= -1.013*10⁶ Pa* (-0.018 m³)
Solving:
W system= 18234 J
<u><em>The amount of work done on the system is 18234 J and the final positive sign means that this work corresponds to an increase in internal energy of the gas.</em></u>
213034 torr is the osmotic pressure.
Explanation:
osmotic pressure is calculated by the formula:
osmotic pressure= iCrT
where i= no. of solute
c= concentration in mol/litre
R= Universal Gas constant
T = temp
It is given that solution is 3% which is 3gms in 100 ml.
let us calculate the concentration in moles/litre
3gm/100ml*1000ml/1L*1mol NaCl/55.84g NaCl
= 5.372 gm/litre
Putting the values in the formula, Temp in Kelvin 318.5K
osmotic pressure= 2*5.372*0.083 * 318.5 Gas constant 0.083
= 284.023 bar or 213018 torr. c= 5.372 moles/L
i=2 for NaCl