Answer:
m = 4450 g
Explanation:
Given data:
Amount of heat added = 4.45 Kcal ( 4.45 kcal ×1000 cal/ 1kcal = 4450 cal)
Initial temperature = 23.0°C
Final temperature = 57.8°C
Specific heat capacity of water = 1 cal/g.°C
Mass of water in gram = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 57.8°C - 23.0°C
ΔT = 34.8°C
4450 cal = m × 1 cal/g.°C × 34.8°C
m = 4450 cal / 1 cal/g
m = 4450 g
The number of significant figures in 369,132,000 is 6
Physical change , the states are changing from liquid to gas
Answer:
373.88 torr
Explanation:
P1 = 350 torr
T1 = 20°C = (20 + 273.15)K = 293.15K
P2 = ?
T2 = 40°C = (40 + 273.15)K = 313.15K
From pressure law,
Pressure of a given mass of gas is directly proportional to its temperature.
P = KT
K = P / T
P1 / T1 = P2 / T2
Solve for P2
P2 = (P1 * T2) / T1
P2 = (350 * 313.15) / 293.15
P2 = 109602.5 / 293.15
P2 = 373.878 torr
P2 = 373.88 torr
The new pressure of the gas would be 373.88 torr.
Answer:
The properties <u>that </u><u>do not vary</u><u> with the variation in the quantity of the material are called as </u><u>intensive property</u>.
While the <u>extensive properties</u><u> are those which </u><u>vary with the variation in the quantity of the material</u>.
Intensive properties:
Mass and volume
Extensive properties:
Density and melting point
Explanation:
The properties <u>that </u><u>do not vary</u><u> with the variation in the quantity of the material are called as </u><u>intensive property</u>.
While the <u>extensive properties</u><u> are those which </u><u>vary with the variation in the quantity of the material</u>.
Intensive properties:
Mass and volume, as with increase in quantity of the material mass and volume increases.
Extensive properties:
Density and melting point, as they remain constant for a object or material
