What mass of water can be heated from 25.0° C to 50.0° C by the addition of 2825 J of heat energy if the specific heat capacity
1 answer:
Answer:
Mass, m = 27g
Explanation:
<u>Given the following data;</u>
Initial Temperature, T1 = 25°C
Final temperature, T2 = 50°C
Quantity of heat = 2825J
Specific heat capacity of water = 4.184
Heat capacity is given by the formula;
Where,
- Q represents the heat capacity.
- m represents the mass of an object.
- c represents the specific heat capacity of water.
- dt represents the change in temperature.
Making mass, m the subject of formula, we have;
Change in temperature, dt = T2 - T1
Change in temperature, dt = 50-25 = 25°C
Substituting into the equation, we have;
Mass, m = 27g
Therefore, the mass of water that can be added is 27 grams.
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Answer:
K.E₂ = mg(h - 2R)
Explanation:
The diagram of the car at the top of the loop is given below. Considering the initial position of the car and the final position as the top of the loop. We apply law of conservation of energy:
K.E₁ + P.E₁ = K.E₂ + P.E₂
where,
K.E₁ = Initial Kinetic Energy = (1/2)mv² = (1/2)m(0 m/s)² = 0 (car initially at rest)
P.E₁ = Initial Potential Energy = mgh
K.E₂ = Final Kinetic Energy at the top of the loop = ?
P.E₂ = Final Potential Energy = mg(2R) (since, the height at top of loop is 2R)
Therefore,
0 + mgh = K.E₂ + mg(2R)
<u>K.E₂ = mg(h - 2R)</u>
