when 84 joules of heat are added to 2.0 grams of water at 15 degrees C, what will be the final temperature of the water?
A. 50. degrees C
B. 5.0 degrees C
C. 25 degrees C
D. 15 degrees C
Let's see if you can do that in your head, knowing that 4.18 joules is 1 calorie. (a good number to remember)
80 joules is therefore very close to 20 calories.
Since the specific heat of water is 1.0 calories per degree-gram, 20 calories is enough to heat 2 grams 10 degrees C.
If you start at 15 C, and add 10, what do you end up with for the temperature?
Well, if I were to summarize this situation in a joke, I'd say that adding heat to water is like giving it a hot beverage - it's bound to warm up! Now, to answer your question, we can use the equation Q = mcΔT, where Q is the heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
In this case, the heat added (Q) is 84 joules, the mass (m) is 2.0 grams, and the initial temperature (ΔT) is 15 degrees Celsius. However, we don't know the specific heat capacity (c) of water.
So, without all the necessary information, I'm afraid I can't give you a definitive answer. But hey, the water is bound to get warmer, right? That's always a safe bet!
To calculate the final temperature of the water, we can use the equation:
Q = m × C × ΔT
where:
Q is the heat energy transferred (in Joules),
m is the mass of the water (in grams),
C is the specific heat capacity of water (4.184 J/g°C), and
ΔT is the change in temperature (in °C).
First, we need to rearrange the equation to solve for ΔT:
ΔT = Q / (m × C)
Given:
Q = 84 J
m = 2.0 g
C = 4.184 J/g°C
Substituting the given values into the equation:
ΔT = 84 J / (2.0 g × 4.184 J/g°C)
ΔT ≈ 10.04 °C
To find the final temperature, we need to add the change in temperature (ΔT) to the initial temperature of the water (15 °C):
Final temperature = Initial temperature + ΔT
Final temperature = 15 °C + 10.04 °C
Final temperature ≈ 25.04 °C
Therefore, the final temperature of the water will be approximately 25.04 degrees Celsius.
The correct answer is C. 25 degrees C.
To find the final temperature of the water, we can use the formula:
Q = m * C * ΔT
Where:
Q is the heat energy input (in joules),
m is the mass of the substance (in grams),
C is the specific heat capacity of the substance (in joules per gram per degree Celsius), and
ΔT is the change in temperature (in degrees Celsius).
First, we need to calculate the heat energy input Q using the given information: Q = 84 joules.
Next, we know the mass of water is 2.0 grams and the initial temperature is 15 degrees Celsius.
Lastly, we need to know the specific heat capacity of water, which is approximately 4.18 joules per gram per degree Celsius.
Now, we can solve for the change in temperature ΔT:
ΔT = Q / (m * C)
ΔT = 84 J / (2.0 g * 4.18 J/g°C)
ΔT ≈ 10.05 °C
To find the final temperature, we add the change in temperature to the initial temperature:
Final temperature = Initial temperature + ΔT
Final temperature = 15 °C + 10.05 °C
Final temperature ≈ 25.05 °C
Therefore, the answer is C. The final temperature of the water will be approximately 25 degrees Celsius.