How much energy is required to raise the temperature of a 300.0
1 answer:
Answer:
Q = 1455.12 Joules.
Explanation:
Given the following data;
Mass = 300 grams
Initial temperature = 22.3
Final temperature = 59.9°C
Specific heat capacity = 0.129 J/gºC.
To find the quantity of energy;
Where,
Q represents the heat capacity.
m represents the mass of an object.
c represents the specific heat capacity of water.
dt represents the change in temperature.
dt represents the change in temperature.
dt = T2 - T1
dt = 59.9 - 22.3
dt = 37.6°C
Substituting the values into the equation, we have;
Q = 1455.12 Joules.
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This is a straightforward dilution calculation that can be done using the equation
where <em>M</em>₁ and <em>M</em>₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and <em>V</em>₁ and <em>V</em>₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.
Here, we have the initial concentration (<em>M</em>₁) and the initial (<em>V</em>₁) and final (<em>V</em>₂) volumes, and we want to find the final concentration (<em>M</em>₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for <em>M</em>₂:
Substituting in our values, we get
So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.