pls help.
1. Calculate the amount of heat necessary to raise the temperature of a 3 kg sample of aluminum from 40°C to 95°C if the specific heat capacity is 900 J/kg°C.
2. Copper has a specific heat capacity of 385 J/kg°C. What is the temperature change of a 4.1 kg sample of copper when 780 J of energy is applied?
3. A 1.1 kg piece of iron absorbs 15686 J of energy when the temperature changes from 16°C to 47°C. What is the specific heat capacity of iron?
4. How much heat is removed to lower the temperature of a sample of a 0.778 kg sample of water from 94°C to 26°C if the specific heat capacity of water is 4186 J/kg°C?
5. You are given three metal samples and you apply the same amount of heat to each one. The temperature changes of the samples vary as follows: Sample 1 changes 20°C, Sample 2 changes 35°C, and Sample 3 changes 50°C. Which sample has the highest specific heat capacity and why?
Sample 3 has the highest specific heat capacity because it requires the most heat to cause a temperature change of the same magnitude.
Sure, I can help you with these questions. These questions involve calculating the amount of heat required or specific heat capacity of different substances. To solve these problems, we will need to use the formula:
Q = mcΔT
Where:
Q is the amount of heat in joules
m is the mass of the substance in kilograms
c is the specific heat capacity of the substance in joules per kilogram per degree Celsius (J/kg°C)
ΔT is the change in temperature in degrees Celsius.
Let's go through each question step by step:
1. To calculate the amount of heat necessary to raise the temperature, we will use the formula mentioned above. We are given:
m = 3 kg (mass of aluminum)
c = 900 J/kg°C (specific heat capacity of aluminum)
ΔT = 95°C - 40°C = 55°C (change in temperature)
Substituting the values into the formula, we get:
Q = (3 kg)(900 J/kg°C)(55°C) = 148,500 J
Therefore, the amount of heat necessary to raise the temperature of the aluminum sample from 40°C to 95°C is 148,500 J.
2. To calculate the temperature change, we will rearrange the formula to solve for ΔT:
ΔT = Q / (mc)
We are given:
m = 4.1 kg (mass of copper)
c = 385 J/kg°C (specific heat capacity of copper)
Q = 780 J (amount of energy applied)
Substituting the values into the formula, we get:
ΔT = 780 J / (4.1 kg)(385 J/kg°C) ≈ 0.49°C (rounded to two decimal places)
Therefore, the temperature change of the copper sample is approximately 0.49°C when 780 J of energy is applied.
3. To calculate the specific heat capacity, we can rearrange the formula to solve for c:
c = Q / (mΔT)
We are given:
m = 1.1 kg (mass of iron)
ΔT = 47°C - 16°C = 31°C (change in temperature)
Q = 15686 J (amount of energy absorbed)
Substituting the values into the formula, we get:
c = 15686 J / (1.1 kg)(31°C) ≈ 450 J/kg°C (rounded to the nearest whole number)
Therefore, the specific heat capacity of iron is approximately 450 J/kg°C.
4. To calculate the heat removed, we will use the same formula mentioned above:
Q = mcΔT
We are given:
m = 0.778 kg (mass of water)
c = 4186 J/kg°C (specific heat capacity of water)
ΔT = 26°C - 94°C = -68°C (negative because the temperature is decreasing)
Substituting the values into the formula, we get:
Q = (0.778 kg)(4186 J/kg°C)(-68°C) ≈ -203,971 J (negative sign indicates heat removed)
Therefore, the heat removed to lower the temperature of the water sample from 94°C to 26°C is approximately 203,971 J.
5. In this question, we have to determine the sample with the highest specific heat capacity based on the temperature changes when the same amount of heat is applied.
For different substances with the same amount of heat applied, the temperature change is inversely proportional to the specific heat capacity. This means that the substance with the highest specific heat capacity will undergo the smallest temperature change.
In this case, we are given three temperature changes:
Sample 1: ΔT1 = 20°C
Sample 2: ΔT2 = 35°C
Sample 3: ΔT3 = 50°C
From these values, we can conclude that the sample with the highest temperature change (ΔT3 = 50°C) has the lowest specific heat capacity since it required the most heat to achieve the temperature change. Conversely, the sample with the smallest temperature change (ΔT1 = 20°C) has the highest specific heat capacity since it required the least heat to achieve the temperature change.
Therefore, Sample 1 has the highest specific heat capacity because it had the smallest temperature change when the same amount of heat was applied.
I hope this explanation helps! Let me know if you have any further questions.
Sure, I can help you with these questions step-by-step.
1. To calculate the amount of heat necessary, you can use the formula: Q = m * c * ΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the temperature change.
Given:
m = 3 kg
c = 900 J/kg°C
ΔT = (95 - 40)°C = 55°C
Now, substitute these values into the formula:
Q = 3 kg * 900 J/kg°C * 55°C
Calculating:
Q = 148,500 J
Therefore, the amount of heat necessary to raise the temperature of the aluminum sample from 40°C to 95°C is 148,500 J.
2. Similar to the previous question, you can use the same formula: Q = m * c * ΔT.
Given:
m = 4.1 kg
c = 385 J/kg°C
Q = 780 J
Rearranging the formula to solve for ΔT:
ΔT = Q / (m * c)
Substituting the given values:
ΔT = 780 J / (4.1 kg * 385 J/kg°C)
Calculating:
ΔT ≈ 0.49°C
Therefore, the temperature change of the copper sample when 780 J of energy is applied is approximately 0.49°C.
3. To find the specific heat capacity of iron, we can rearrange the formula Q = m * c * ΔT to solve for c.
Given:
m = 1.1 kg
Q = 15686 J
ΔT = (47 - 16)°C = 31°C
Rearranging the formula:
c = Q / (m * ΔT)
Substituting the given values:
c = 15686 J / (1.1 kg * 31°C)
Calculating:
c ≈ 470 J/kg°C
Therefore, the specific heat capacity of iron is approximately 470 J/kg°C.
4. Using the same formula as before, Q = m * c * ΔT, we can calculate the amount of heat removed.
Given:
m = 0.778 kg
c = 4186 J/kg°C
ΔT = (26 - 94)°C = -68°C (negative because the temperature decreased)
Substituting the values into the formula:
Q = 0.778 kg * 4186 J/kg°C * -68°C
Calculating:
Q = -198,593.568 J
Therefore, the amount of heat removed to lower the temperature of the water sample from 94°C to 26°C is approximately -198,593.568 J (negative because the heat was removed).
5. The specific heat capacity of a substance determines how much heat energy it can absorb for a given temperature change. The sample with the highest specific heat capacity would be the one that experiences the smallest temperature change when the same amount of heat energy is applied.
In this case, Sample 1 changes by 20°C, Sample 2 changes by 35°C, and Sample 3 changes by 50°C.
Therefore, Sample 1 has the highest specific heat capacity because it experienced the smallest temperature change, indicating that it can absorb or release heat energy more efficiently than the other samples.