What is the amount of heat (in Joules) absorbed by 18 g of
water initially at room temperature heated to 100°C? If 10g
of Cu is added to this water , than decrease in temperature
in Kelvin) of water was found to be? C(p,m) for water
75.32J/mol K; C(p,m) for Cu = 24.47J/mol K.
Answers
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Explanation:
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Answer:
The following formula may determine how much heat 18 grams of water at room temperature (assumed to be 25 degrees Celsius) absorbed when heated to 100 degrees Celsius.
Q = mcΔT
Explanation:
Using the formula Q = m * c + T, where m is the mass of water, c is its specific heat capacity, and T is the temperature difference.
We also know that the specific heat capacity of water is 75.32 J/mol K, and its mass is 18 g. Taking the difference between the starting and ending temperatures gives us T.
ΔT = 100°C - 25°C
ΔT = 75°C
These numbers may now be used in the formula:
Q = mcΔT
Quantity (Q) = Mass (g) × Energy Per Mole (J/K) x Temperature (°C + 273.15). K
Q = 1.35 x 10^4 J
Hence, 1.35 x 104 Joules is the quantity of heat the water absorbs.
Copper, We need to determine how much 10 g of Copper will raise the water temperature next. Consider this helpful formula:
Q = mcΔT
Water heat loss is denoted by Q, where m is the total mass of water and Cu, c is the specific heat capacity of the water-and-Cu combination, and T is the temperature difference.
For this calculation, we used the specific heat capacity of Cu, which we know to be 24.47 J/mol K, and the masses of the water and Cu, which we know to be 18 g + 10 g = 28 g. To determine the mixture's specific heat capacity, we may use the following formula:
Mixing Concentration (Cp, mix) = (Cp, water x water + Cp, Cu x mCu) / (water + mCu)., where Cp, water is water's specific heat capacity, water is water's mass, Cp, Cu is Copper's specific heat capacity, and mCu is Copper's mass.
With the introduction of the values, we obtain:
Cp,mix = (75.32 J/mol K x 18 g + 24.47 J/mol K x 10 g) / 28 g Cp,mix = 63.32 J/mol K
Now we can plug in the values into the formula for Q:
We use the formula Q = mcT -1.35 x 104 J = 28 g x 63.32 J/mol K x T to calculate the heat capacity.
Finding T's solution yields
ΔT = -6.02 K
Therefore, the temperature of the water decreases by 6.02 Kelvin when 10 g of Cu is added.
To learn more about heat, click on the link below
https://brainly.in/question/1567286
To learn more about room temperature, click on the link below
https://brainly.in/question/91518
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