Answer:
48.37514 kj
Explanation:
Given data:
Mass of water = 163 g
Initial temperature = 29°C
Final temperature = 100°C
Heat added = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Specific heat capacity of water is 4.18 j/g.°C
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 = Final temperature - initial temperature
ΔT = 100°C - 29°C
ΔT = 71°C
Q = 163 g × 4.18 j/g.°C × 71°C
Q = 48375.14 j
Joule to Kj conversion:
48375.14 /1000 = 48.37514 kj
Answer:
Hope this is helpful to you!
The independent variable is the one that is changed by the scientist. To insure a fair test, a good experiment has only ONE independent variable. As the scientist changes the independent variable, he or she records the data that they collect.
Aluminum or glass I think
Answer:
V = 364500 L, 476.748 yard³
Explanation:
Given that,
The dimensions of a room are 10 meters wide by 15 meters long and 8.0 ft high.
l = 10 m, b = 15 m, h = 8 ft = 2.43 m
The volume of the room is :
V = lbh
So,
V = 10×15×2.43
V = 364.5 m³
As 1 m³ = 1000 L
364.5 m³ = 364500 L
Also, 1 m³ = 1.30795 yard³
364.5 m³ = 476.748 yard³
Hence, this is the required solution.
