1050 calories of heat from soybean oil was applied to a 250 gram sample of liquid water with an initial temperature of 25 degrees Celsius. Determine (a) the final temperature of water and (b) the change in temperature.
To determine the final temperature of the water and the change in temperature, we can use the equation:
Q = m * c * ΔT
where Q is the amount of heat transferred, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
Given:
Q = 1050 calories
m = 250 grams
Initial temperature = 25 degrees Celsius
(a) To find the final temperature of the water, we use the equation:
Q = m * c * ΔT
Rearranging the equation, we get:
ΔT = Q / (m * c)
The specific heat capacity of water is approximately 1 calorie/gram•degree Celsius.
Substituting the values into the equation:
ΔT = 1050 calories / (250 grams * 1 calorie/gram•degree Celsius)
ΔT = 1050 calories / 250 grams
ΔT ≈ 4.2 degrees Celsius
Therefore, the change in temperature is approximately 4.2 degrees Celsius.
(b) To find the final temperature of the water, we add the change in temperature to the initial temperature:
Final temperature = Initial temperature + ΔT
Final temperature = 25 degrees Celsius + 4.2 degrees Celsius
Final temperature ≈ 29.2 degrees Celsius
Therefore, the final temperature of the water is approximately 29.2 degrees Celsius.
To solve this problem, we need to apply the concept of specific heat capacity, which is the amount of heat required to raise the temperature of a substance by 1 degree Celsius per gram. For water, the specific heat capacity is 4.18 J/g°C.
(a) To find the final temperature of the water, we need to calculate the heat transferred from the soybean oil to the water. We can use the formula:
Q = mcΔT
where Q is the heat transferred, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
Given:
Q = 1050 calories
m (mass of water) = 250 grams
c (specific heat capacity of water) = 4.18 J/g°C
ΔT (change in temperature) = final temperature - initial temperature
First, we need to convert the heat from calories to joules, as the specific heat capacity is in joules. To do this, we use the conversion factor: 1 calorie = 4.18 joules.
So, the heat transferred (Q) is:
Q = 1050 calories × 4.18 J/calorie = 4389 J
Next, let's plug the values into the formula and solve for the final temperature (Tf):
4389 J = 250 g × 4.18 J/g°C × (Tf - 25°C)
Simplifying the equation:
4389 J = 1045 g × (Tf - 25°C)
Dividing both sides by 1045 g:
4.2 J/g = Tf - 25°C
Adding 25°C to both sides:
Tf = 4.2 J/g + 25°C
Therefore, the final temperature of the water is approximately 29.2°C.
(b) To find the change in temperature, we subtract the initial temperature from the final temperature:
Change in temperature = Final temperature - Initial temperature
Change in temperature = 29.2°C - 25°C
Change in temperature = 4.2°C
Thus, the change in temperature is approximately 4.2°C.