Answer:
The final temperature is:- 7428571463.57 °C
Explanation:
The expression for the calculation of heat is shown below as:-
Where,
is the heat absorbed/released
m is the mass
C is the specific heat capacity
is the temperature change
Thus, given that:-
Mass of water = 1.75 mg = 0.00175 g ( 1 g = 0.001 mg)
Specific heat of water = 4.18 J/g°C
Initial temperature = 35 °C
Final temperature = x °C
![\Delta T=(x-35)\ ^0C/tex] Q = [tex]1.3\times 10^4](https://tex.z-dn.net/?f=%5CDelta%20T%3D%28x-35%29%5C%20%5E0C%2Ftex%5D%0A%3C%2Fp%3E%3Cp%3EQ%20%3D%20%5Btex%5D1.3%5Ctimes%2010%5E4)
kcal
Also, 1 kcal = 4.18 kJ = 
J
So, Q = 
J = 54340000 J
So,
Thus, the final temperature is:- 7428571463.57 °C
Dangerous
hope i helpedXD
The answer is A because it’s how you calculate the mass
Answer:
The Kinetic Energy is approximately 3 times decreased
Explanation:
A baseball weighs 5.13 oz.
a)What is the kinetic energy, in joules, of this baseball when it is thrown by a major league pitcher at 95.o mi/h?
b) By what factor will the kinetic energy change if the speed of the baseball is decreased to 54.8 mi/h? Express your answer as an integer.
Kinetic Energy (KE)=0.5×mass×velocity ^ 2
Kinetic Energy (KE)=0.5×mass × velocity ^ 2
Joules = kg×m^2/s^2
1 mile = 1609.344 meters
1 hour = 3600 sec
1 Oz = 28.34952 g = 0.02834952 kg
a) KE=0.5×m×v^2
=0.5×(5.13 oz × 0.02834952 kg/1 ounce)×(95 miles/h × 1609.344 m/1 mile × 1 hr/3600 s)^2
=130.761 kg×m^2/s^2 = 130.761 Joules
b) KE=0.5×m×v^2
=0.5×(5.13 oz × 0.02834952 kg/1 ounce)×(54.8 miles/h × 1609.344 m/1 mile × 1 hr/3600 s)^2
=43.51028 kg×m^2/s^2 = 43.51028 Joules
= 130.761 / 43.51028 = 3.00528,
As such the Kinetic Energy is approximately 3 times decreased
Answer:
Temperature stays constant.
Explanation:
During a phase change, the temperature does not change. Temperature will only change when a phase change is completed.
