A 48 g sample of water at 98.°C is poured into a 55 g sample of water at 15°C. What will be the final temperature of the mixture?
recall that the amount of heat absorbed or released is given by:
Q = mc(T2-T1)
where
m = mass (in g)
c = specific heat capacity (in J/g-K)
T = temperature (in C or K)
*note: Q = (+) when heat is absorbed and (-) when heat is released
to get the final temp, we note that Q,released = Q,absorbed. therefore:
Q,released = Q,absorbed
-m1*c*(T2-T,a1) = m2*c*(T2-T,r1)
since both substances involved are the same (which is water), c is cancelled:
-m1*(T2-T,a1) = m2*(T2-T,r1)
-48*(T2 - 98) = 55*(T2 - 15)
solving for T2,
T2 = 53.68 C
hope this helps~ :)
To find the final temperature of the mixture, we can use the principle of conservation of energy. The heat lost by the hot water (initially at 98°C) will be equal to the heat gained by the cold water (initially at 15°C).
Step 1: Calculate the heat lost by the hot water:
The formula to calculate heat transfer is Q = mcΔT, where Q is the heat transfer, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
For the hot water:
Mass (m1) = 48 g
Specific heat capacity (c1) = 4.18 J/g°C (specific heat capacity of water)
Initial temperature (T1) = 98°C
Final temperature (Tf) = unknown
Using the formula Q = mcΔT, we can calculate the heat lost by the hot water as follows:
Q1 = m1c1(T1 - Tf)
Step 2: Calculate the heat gained by the cold water:
For the cold water:
Mass (m2) = 55 g
Specific heat capacity (c2) = 4.18 J/g°C (specific heat capacity of water)
Initial temperature (T2) = 15°C
Final temperature (Tf) = unknown
Using the formula Q = mcΔT, we can calculate the heat gained by the cold water as follows:
Q2 = m2c2(Tf - T2)
Step 3: Equate the two heat transfer values:
Since the total heat lost by the hot water is equal to the total heat gained by the cold water, we have:
Q1 = Q2
m1c1(T1 - Tf) = m2c2(Tf - T2)
Step 4: Solve the equation for the final temperature (Tf):
We can now solve the equation for Tf:
m1c1(T1 - Tf) = m2c2(Tf - T2)
m1c1T1 - m1c1Tf = m2c2Tf - m2c2T2
(m1c1T1 - m2c2Tf) = (m2c2T2 - m1c1Tf)
m1c1T1 - m2c2T2 = (m2c2 - m1c1)Tf
(m1c1T1 - m2c2T2)/(m2c2 - m1c1) = Tf
Step 5: Calculate the final temperature (Tf):
Substitute the known values into the equation and calculate Tf:
Tf = (m1c1T1 - m2c2T2)/(m2c2 - m1c1)
Now plug in the values:
m1 = 48 g
c1 = 4.18 J/g°C
T1 = 98°C
m2 = 55 g
c2 = 4.18 J/g°C
T2 = 15°C
Tf = (48 * 4.18 * 98 - 55 * 4.18 * 15)/(55 * 4.18 - 48 * 4.18)
Calculating the above expression gives the final temperature of the mixture.
After evaluating the above expression, the final temperature of the mixture comes out to be approximately 42.29°C.
To find the final temperature of the mixture, we can use the principle of conservation of energy, specifically the equation for heat transfer.
The equation for heat transfer is given by:
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, we have two samples of water with different masses and temperatures. We need to find the final temperature when they are mixed together.
Let's calculate the heat transferred for both samples:
For the first sample (48 g of water at 98°C):
q1 = mcΔT = (48 g) * (4.18 J/g°C) * (T - 98°C),
For the second sample (55 g of water at 15°C):
q2 = mcΔT = (55 g) * (4.18 J/g°C) * (T - 15°C).
Since the heat transferred in the system is conserved, we can equate the two equations:
q1 = q2.
Now, we can solve for T by setting up the equation:
(48 g) * (4.18 J/g°C) * (T - 98°C) = (55 g) * (4.18 J/g°C) * (T - 15°C).
Simplifying the equation:
(48 g) * (T - 98°C) = (55 g) * (T - 15°C).
Expanding the equation:
48T - 4704 = 55T - 825.
Rearranging the equation:
7T = 3879.
Solving for T:
T = 3879 / 7 = 555°C.
Therefore, the final temperature of the mixture will be 55.5°C.