4. A 70.0 g sample of water at 95.00°C is mixed with 50.0 g of water at 135.0°C. Calculate the final temperature of the mixture assuming no heat loss to the surroundings. Remember that the heat gained by the cooler water is equal to the heat lost by the warmer water.
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To calculate the final temperature of the mixture, we can use the principle of conservation of energy. As stated in the question, we assume no heat loss to the surroundings, so the energy gained by the cooler water is equal to the energy lost by the warmer water.
The energy gained or lost by a substance can be calculated using the following equation:
Q = mcΔT
Where:
Q is the energy gained or lost
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature
First, let's find the amount of energy gained by the cooler water. We can assign this as positive since it is gaining energy:
Q1 = mcΔT1
Where:
m1 = 70.0 g (mass of cooler water)
c1 = specific heat capacity of water = 4.18 J/g°C (assume constant for both waters)
ΔT1 = final temperature - initial temperature of cooler water
Next, let's find the energy lost by the warmer water. We assign this as negative since it is losing energy:
Q2 = -mcΔT2
Where:
m2 = 50.0 g (mass of warmer water)
c2 = specific heat capacity of water = 4.18 J/g°C (assume constant for both waters)
ΔT2 = initial temperature of warmer water - final temperature
Since the energy gained by the cooler water is equal to the energy lost by the warmer water, we have:
Q1 = Q2
Therefore:
m1c1ΔT1 = -m2c2ΔT2
Now let's plug in the given values:
70.0 g * 4.18 J/g°C * (final temperature - 95.00°C) = -50.0 g * 4.18 J/g°C * (135.0°C - final temperature)
Simplifying this equation will lead us to the final temperature of the mixture. I'll calculate that for you.