139 g of water at 4.9 degrees Celsius mixed with 241 g of water at 96.0 degrees Celsius (Tf=62.7)
some how the question is missing
To solve this problem, we can use the heat transfer equation:
Q = m × c × ∆T
where:
Q = heat transferred (in joules)
m = mass (in grams)
c = specific heat capacity of water (4.18 J/g°C)
∆T = change in temperature (in °C)
Let's calculate the heat transferred between the two water samples.
For the first water sample:
m1 = 139 g
T1 = 4.9°C
Tf = 62.7°C
∆T1 = Tf - T1 = 62.7°C - 4.9°C = 57.8°C
Q1 = m1 × c × ∆T1 = 139 g × 4.18 J/g°C × 57.8°C = 33789.4 J
For the second water sample:
m2 = 241 g
T2 = 96.0°C
Tf = 62.7°C
∆T2 = Tf - T2 = 62.7°C - 96.0°C = -33.3°C
Q2 = m2 × c × ∆T2 = 241 g × 4.18 J/g°C × (-33.3°C) = -33167.454 J
Since heat is transferred from the second water sample to the first, we need to consider the negative sign.
Now, let's find the final temperature when the two water samples are mixed.
Q1 + Q2 = 0
33789.4 J - 33167.454 J = 0
33789.4 J = 33167.454 J
Since the heat lost by the second water sample is equal to the heat gained by the first water sample, we can equate the absolute values:
33789.4 J = 33167.454 J
Therefore, the final mixture temperature (Tf) is 62.7°C.
To solve this problem, you can use the principle of conservation of energy. The heat lost by the hot water is equal to the heat gained by the cold water.
First, let's calculate the amount of heat lost by the hot water. We can use the formula:
Q_hot = m_hot * c * (T_hot - Tf)
Where:
Q_hot is the heat lost by the hot water
m_hot is the mass of the hot water (241 g)
c is the specific heat capacity of water (4.18 J/g°C)
T_hot is the initial temperature of the hot water (96.0°C)
Tf is the final temperature of both the hot and cold water (62.7°C)
Plugging in the values:
Q_hot = 241 g * 4.18 J/g°C * (96.0°C - 62.7°C)
Next, let's calculate the amount of heat gained by the cold water. We can use the formula:
Q_cold = m_cold * c * (Tf - T_cold)
Where:
Q_cold is the heat gained by the cold water
m_cold is the mass of the cold water (139 g)
c is the specific heat capacity of water (4.18 J/g°C)
Tf is the final temperature of both the hot and cold water (62.7°C)
T_cold is the initial temperature of the cold water (4.9°C)
Plugging in the values:
Q_cold = 139 g * 4.18 J/g°C * (62.7°C - 4.9°C)
Now, according to the principle of conservation of energy, Q_hot = Q_cold. Therefore:
241 g * 4.18 J/g°C * (96.0°C - 62.7°C) = 139 g * 4.18 J/g°C * (62.7°C - 4.9°C)
Solving for Tf, the final temperature:
(241 g * 4.18 J/g°C * (96.0°C - 62.7°C)) / (139 g * 4.18 J/g°C) + 4.9°C = Tf
Evaluating this expression will give you the final temperature at which both the hot and cold water mix.