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?
1. To calculate the amount of heat necessary to raise the temperature of a sample, we can use the formula:
Q = m * c * ΔT
where:
Q = amount of heat (in joules)
m = mass of the sample (in kilograms)
c = specific heat capacity (in joules per kilogram per degree Celsius)
ΔT = change in temperature (in degrees Celsius)
In this case, we have:
m = 3 kg
c = 900 J/kg°C
ΔT = (95°C - 40°C) = 55°C
Plugging these values into the formula, we get:
Q = 3 kg * 900 J/kg°C * 55°C
Q = 148,500 J
Therefore, the amount of heat necessary to raise the temperature of the 3 kg aluminum sample from 40°C to 95°C is 148,500 joules.
2. Using the same formula as before:
Q = m * c * ΔT
In this case, we have:
m = 4.1 kg
c = 385 J/kg°C
Q = 780 J
We need to find ΔT in this case. Rearranging the formula:
ΔT = Q / (m * c)
ΔT = 780 J / (4.1 kg * 385 J/kg°C)
ΔT ≈ 0.48°C
Therefore, the temperature change of the 4.1 kg copper sample when 780 J of energy is applied is approximately 0.48°C.
3. To find the specific heat capacity (c) of iron, we can rearrange the formula:
Q = m * c * ΔT
In this case, we have:
m = 1.1 kg
Q = 15,686 J
ΔT = (47°C - 16°C) = 31°C
Rearranging the formula:
c = Q / (m * ΔT)
c = 15,686 J / (1.1 kg * 31°C)
c ≈ 500 J/kg°C
Therefore, the specific heat capacity of iron is approximately 500 J/kg°C.
4. We can use the same formula:
Q = m * c * ΔT
In this case, we have:
m = 0.778 kg
c = 4186 J/kg°C
ΔT = (26°C - 94°C) = -68°C (negative because the temperature is decreasing)
Plugging in the values:
Q = 0.778 kg * 4186 J/kg°C * (-68°C)
Q = -201,461 J
Therefore, the amount of heat removed to lower the temperature of the 0.778 kg water sample from 94°C to 26°C is approximately 201,461 joules.
5. To determine which sample has the highest specific heat capacity, we need to compare the temperature changes and see which one requires the most heat input per degree Celsius increase.
Sample 1: ΔT = 20°C
Sample 2: ΔT = 35°C
Sample 3: ΔT = 50°C
Since the same amount of heat is applied to each sample, the sample with the highest temperature change (ΔT) would indicate higher specific heat capacity, as it required more heat to achieve that change in temperature. Therefore, Sample 3, which changes by 50°C, has the highest specific heat capacity among the three samples.