You take 326 g of a solid (melting point = 57.6oC, enthalpy of fusion = 346 J/g) and let it melt in 757
g of water, and the water temperature decreases from its initial temperature to 57.6oC. Calculate the
initial temperature of the water. (Note that the specific heat capacity of water is 4.18 J/(goC).)
[mass solid x heat fusion] + [mass water x specific heat water x (Tf-Ti)] = 0
Substitute and solve for Ti.
Well, this sounds like a melting hot situation! Let's break it down.
First, let's determine the heat change when the solid melts. The formula for heat change is Q = m * Hf, where Q is the heat change, m is the mass, and Hf is the enthalpy of fusion. So, the heat change when the solid melts is Q = 326 g * 346 J/g = 112796 J.
Next, let's calculate the heat change when the water cools down. The formula for heat change is Q = m * c * ΔT, where Q is the heat change, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. We know the initial temperature of water is Ti, its mass is 757 g, and the specific heat capacity of water is 4.18 J/(g°C). The change in temperature is Ti - 57.6°C.
Since energy is conserved, the heat change when the solid melts must be equal to the heat change when the water cools down. Therefore, we can set up the equation:
112796 J = 757 g * 4.18 J/(g°C) * (Ti - 57.6°C)
Now, we can solve for Ti:
112796 J = 3161.26 g°C * Ti - 183195.36 J
Rearranging the equation, we have:
3161.26 g°C * Ti = 112796 J + 183195.36 J
Ti = (112796 J + 183195.36 J) / 3161.26 g°C
Calculating the sum, we find:
Ti ≈ 89.25°C
So, the initial temperature of the water is approximately 89.25°C. Now that's hot enough to make a cup of tea!
To find the initial temperature of the water, we can use the concept of heat transfer.
The heat lost by the solid (Q1) is equal to the heat gained by the water (Q2):
Q1 = Q2
The heat lost by the solid can be calculated using the formula:
Q1 = mass of solid × enthalpy of fusion
Given that the mass of the solid is 326 g and the enthalpy of fusion is 346 J/g, we can substitute these values into the formula:
Q1 = 326 g × 346 J/g
Q1 = 112796 J
The heat gained by the water can be calculated using the formula:
Q2 = mass of water × specific heat capacity of water × temperature change
The mass of water is given as 757 g, and the specific heat capacity of water is 4.18 J/(g°C). The temperature change is the difference between the initial temperature of the water and the melting point of the solid, which is 57.6°C.
Q2 = 757 g × 4.18 J/(g°C) × (initial temperature - 57.6°C)
We can now substitute the values into the formula and set the equations equal to each other:
112796 J = 757 g × 4.18 J/(g°C) × (initial temperature - 57.6°C)
Now we can solve for the initial temperature of the water:
112796 J = 3161.26 g × (initial temperature - 57.6°C)
(initial temperature - 57.6°C) = 112796 J / 3161.26 g
(initial temperature - 57.6°C) = 35.67°C
initial temperature = 35.67°C + 57.6°C
initial temperature = 93.27°C
Therefore, the initial temperature of the water is 93.27°C.
To solve this problem, we can use the concept of heat transfer.
The heat gained by the solid when it melts is equal to the heat lost by the water when it cools down. The heat gained by the solid is given by:
Heat gained by the solid = Mass of the solid x Enthalpy of fusion
Heat gained by the solid = 326 g x 346 J/g = 112796 J
The heat lost by the water can be calculated using the following formula:
Heat lost by the water = Mass of the water x Specific heat capacity of water x Change in temperature
Here, we want to find the initial temperature of the water, so let's call it T.
The change in temperature for the water is given by the difference between the initial temperature of the water and the melting point of the solid, which is 57.6oC.
Change in temperature = T - 57.6oC
Now, we can substitute these values into the formula for heat lost by the water:
112796 J = 757 g x 4.18 J/(g⋅oC) x (T - 57.6oC)
To solve for T, let's rearrange the equation:
112796 J = 3160.26(T - 57.6oC)
Divide both sides of the equation by 3160.26:
T - 57.6oC = 112796 J / 3160.26
T - 57.6oC ≈ 35.70
Now, add 57.6oC to both sides of the equation:
T = 35.70 + 57.6oC
T ≈ 93.30oC
Therefore, the initial temperature of the water is approximately 93.30oC.