Answer:
I believe this is a K-12 test question. If the answers below are what you have on your test . . .
- Precise
- Accurate
- Identical
- None of the above
Then the answer is <u>precise</u>.
Answer:
The final temperature of the system is 27.3°C.
Explanation:
Heat lost by aluminum = 3.99 × 0.91 × (100-T)
= 3.631 (100-T)
Heat gained by water = 10 × 4.184 × (T-21)
= 41.84 (T-21)
As,
Heat gained = Heat loss
or, 3.631(100-T) = 41.84(T-21)
or,363.1 - 3.631 T = 41.84 T - 878.64)
or, (41.84+ 3.631) T = 878.64 +363.1
or T= 
or, T = 27.3°C
Hence the final temperature is 27.3°C.
Kepler did not study the speed of the planets, rather, he studied how the planets move in the solar system. He proposed three laws. As a summary, he described that the planets move around the sun in the shape of an ellipse (orbit), and the Sun being one of the foci. Then, he proposed the period for the planet to complete one revolution around the Sun.
On the other hand, Newton studied the forces acting on the planet (or any object in space) that explain how the planets move around the solar system as described by Kepler. Also, Kepler's observations only apply to planets and not the moons or satellites. Thus, Kepler only made laws from observations, while Newton based it from underlying principles that led him to mathematical equations such as the law of universal gravitation.
Answer:
5 g / ml
Explanation:
Convert the values given to g and ml
12.5 kg = 12500 g
2.5 L = 2500 ml
12500 g / 2500 ml = 5 g/ml
Answer:
T2 = 135.1°C
Explanation:
Given data:
Mass of water = 96 g
Initial temperature = 113°C
Final temperature = ?
Amount of energy transfer = 1.9 Kj (1.9×1000 = 1900 j)
Specific heat capacity of aluminium = 0.897 j/g.°C
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 = T2 - T1
Now we will put the values in formula.
Q = m.c. ΔT
1900 j = 96 g × 0.897 j/g.°C × T2 - 113°C
1900 j = 86.112 j/°C × T2 - 113°C
1900 j / 86.112 j/°C = T2 - 113°C
22.1°C + 113°C = T2
T2 = 135.1°C
