1. How much heat energy is lost by 3 kg of water when it cools from 80 degrees C to 10 degrees C?
2. A 300 g piece of aluminum is heated from 30 degrees C to 150 degrees C. What amount of heat energy is absorbed?
3. Determine the temperature change in each of the following.
(a) 10 kg of water loses 232 kJ of heat energy.
(b) 500 g of copper gains 1.96 kJ of heat energy.
4. After 2 kg of mercury gained 2.52 x 10^4 J of heat energy, its final temperature was 130 degrees C. What was its initail temperature?
#1=-877,590 J
1. Well, losing heat is never a cool thing, but let's crunch some numbers. The specific heat capacity of water is approximately 4.18 J/g°C. So, for 3 kg of water, the heat energy lost can be calculated using the formula:
Q = mcΔT
where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
ΔT = 80°C - 10°C = 70°C
Q = (3 kg) * (4.18 J/g°C) * (70°C)
And voila! You'll have your answer, I promise it won't be water under the bridge.
2. Wow, that piece of aluminum must have been really hot! To find the amount of heat energy absorbed, we can use the same formula as before:
Q = mcΔT
Given that the specific heat capacity of aluminum is approximately 0.897 J/g°C, we have:
ΔT = 150°C - 30°C = 120°C
Q = (300 g) * (0.897 J/g°C) * (120°C)
And there you have it, the heat energy absorbed by that toasty aluminum.
3. Let's get calculating, temperature change detectives!
(a) This time, we need to rearrange the formula Q = mcΔT to solve for ΔT. Dividing both sides by mc, we get:
ΔT = Q / mc
Given that Q = 232 kJ = 232,000 J, m = 10 kg, and c = 4.18 J/g°C (since it's still water), we have:
ΔT = (232,000 J) / (10 kg * 4.18 J/g°C)
(b) Now, let's use the same formula, but this time with copper as our star. Given that Q = 1.96 kJ = 1,960 J, m = 500 g, and c = 0.386 J/g°C for copper, we get:
ΔT = (1,960 J) / (500 g * 0.386 J/g°C)
Tada! You've uncovered the temperature changes.
4. Well, it seems like mercury is a bit of a drama queen when it comes to heat energy. To find its initial temperature, we can use the formula Q = mcΔT like we did before. Given that Q = 2.52 x 10^4 J, m = 2 kg, and ΔT = 130°C - initial temperature, we can rearrange the formula to solve for the initial temperature:
initial temperature = ΔT + (Q / mc)
Now let's plug in the values we know:
initial temperature = 130°C - (2.52 x 10^4 J) / (2 kg * specific heat capacity of mercury)
And there you go, the mystery of the initial temperature solved.
To calculate the amount of heat energy gained or lost in each of the given scenarios, we can use the formula:
Q = m * c * ΔT
Where:
Q is the heat energy gained or lost (in joules),
m is the mass of the substance (in kilograms or grams),
c is the specific heat capacity of the substance (in joules per kilogram per degree Celsius or joules per gram per degree Celsius),
ΔT is the change in temperature (in degrees Celsius).
Let's address each question separately:
1. To calculate the heat energy lost by 3 kg of water when it cools from 80 degrees C to 10 degrees C, we need to use the specific heat capacity of water, which is approximately 4,186 J/kg°C.
First, we calculate the change in temperature: ΔT = final temperature - initial temperature
ΔT = 10°C - 80°C = -70°C
Now, we use the formula to calculate the heat energy lost:
Q = m * c * ΔT
Q = 3 kg * 4,186 J/kg°C * (-70°C)
Q = -878,340 J
Therefore, 3 kg of water loses 878,340 joules of heat energy when it cools from 80 degrees C to 10 degrees C.
2. To find the amount of heat energy absorbed by a 300 g piece of aluminum when it is heated from 30 degrees C to 150 degrees C, we will use the specific heat capacity of aluminum, which is approximately 900 J/kg°C.
First, we need to convert the mass to kilograms:
300 g = 0.3 kg
Then, we calculate the change in temperature: ΔT = final temperature - initial temperature
ΔT = 150°C - 30°C = 120°C
Now, we can calculate the heat energy absorbed:
Q = m * c * ΔT
Q = 0.3 kg * 900 J/kg°C * 120°C
Q = 32,400 J
Therefore, a 300 g piece of aluminum absorbs 32,400 joules of heat energy when heated from 30 degrees C to 150 degrees C.
3. (a) To determine the temperature change when 10 kg of water loses 232 kJ of heat energy, we know the mass of water and the amount of heat energy lost.
Let's convert the heat energy from kilojoules to joules:
232 kJ = 232,000 J
We can rearrange the formula to find ΔT:
ΔT = Q / (m * c)
ΔT = 232,000 J / (10 kg * 4,186 J/kg°C)
ΔT ≈ 5.54°C
Therefore, 10 kg of water loses 232 kJ of heat energy, resulting in a temperature change of approximately 5.54°C.
(b) For a 500 g copper sample that gains 1.96 kJ of heat energy, we can use the specific heat capacity of copper, which is approximately 385 J/kg°C.
Let's convert the heat energy from kilojoules to joules:
1.96 kJ = 1,960 J
Again, we rearrange the formula to find ΔT:
ΔT = Q / (m * c)
ΔT = 1,960 J / (0.5 kg * 385 J/kg°C)
ΔT ≈ 10.13°C
Therefore, a 500 g copper sample gains 1.96 kJ of heat energy, resulting in a temperature change of approximately 10.13°C.
4. To determine the initial temperature of 2 kg of mercury after gaining 2.52 x 10^4 J of heat energy and reaching a final temperature of 130 degrees C, we need to use the specific heat capacity of mercury, which is approximately 140 J/kg°C.
We can rearrange the formula to find the initial temperature:
Q = m * c * ΔT
Therefore, ΔT = Q / (m * c)
Let's calculate the change in temperature:
ΔT = 2.52 x 10^4 J / (2 kg * 140 J/kg°C)
ΔT ≈ 90.0°C
Now, we can find the initial temperature by subtracting the change in temperature from the final temperature:
Initial temperature = Final temperature - ΔT
Initial temperature = 130°C - 90.0°C
Initial temperature ≈ 40.0°C
Therefore, the initial temperature of the 2 kg of mercury was approximately 40.0 degrees C before gaining 2.52 x 10^4 J of heat energy and reaching a final temperature of 130 degrees C.
recall that heat absorbed or released is given by
Q = mc(T2 - T1)
where
m = mass (in kg)
c = specific heat capacity (in J/kg-K)
T = temperature (in C or K)
*note: Q is (+) when heat is absorbed, and (-) when heat is released.
#1. we just substitute. the c for water is equal to 4.184 J/kg-K (you can google this or look for this in books):
Q = mc(T2-T1)
Q = 3*4.184*(10-80)
Q = -878.64 J
*note that it is negative since it cooled, and thus released heat.
#2,3 and 4: just substitute the given values and solve for the unknown. be careful with units.
hope this helps~ :)