Answer:
Q = 1057.5 [cal]
Explanation:
In order to solve this problem, we must use the following equation of thermal energy.

where:
Q = heat energy [cal]
Cp = specific heat = 0.47 [cal/g*°C]
T_final = final temperature = 32 [°C]
T_initial = initial temperature = 27 [°C]
m = mass of the substance = 450 [g]
Now replacing:
![Q=450*0.47*(32-27)\\Q=1057.5[cal]](https://tex.z-dn.net/?f=Q%3D450%2A0.47%2A%2832-27%29%5C%5CQ%3D1057.5%5Bcal%5D)
Answer:
2.55 × 10³ J =2.55 kJ
Explanation:
Specific heat capacity of ice = 37.8 J / mol °C
Specific heat capacity of water = 76.0 J/ mol °C
Ice at -12 °C is converted to ice at 0 °C by absorbing heat Q₁
Ice at 0°C melts to water at 0 °C. Let Heat absorbed during this phase change be Q₂ .
Let heat absorbed to raise the temperature of water from 0 C to 24°C be Q₃ .
Total heat = Q = Q₁ + Q₂ + Q₃
Q₁ = (37.8 j/mol C )(5.53 g /18.01532 g/ mol )( 0-(-12)) = 139.23749 j
Q₂ =(5.53 g/18.01532 g H₂O / mol ) (6.02 x10³ j) = 1847.905 j
Q₃ = (76 j/mol C) ( (5.53 g/18.01532 g H₂O / mol )(24-0) = 559.8968 j
Total Heat required = Q = 139.23749 j + 1847.905 j + 559.8968 j
= 2547.039 j = 2.55 × 10³ J =2.55 kJ
Answer:
60*12.0= 720 = v/60 * 12.0 squared which is 1,728
Explanation:
Horizontal velocity component: Vx = V * cos(α)
Answer:

Explanation:
Work is the product of force and distance.

We know that 96 Joules of work were done and a 16 Newton force was applied to the object.
Substitute the values into the formula.

First, let's convert the units. This will make cancelling units easier later in the problem. 1 Joule (J) is equal to 1 Newton meter (N*m), so the work of 96 Joules equals 96 Newton meters.

Now, solve for distance by isolating the variable, d. It is being multiplied by 16 Newtons and the inverse of multiplication is division. Divide both sides of the equation by 16 N.


The units of Newtons cancel.


The object moved a distance of <u>6 meters.</u>