Logo

AskSia

Plus

A 22 g block of copper at 45 °C absorbs 2.59 kJ of heat. Given the spefific heat...
Oct 9, 2024
A 22 g block of copper at 45 °C absorbs 2.59 kJ of heat. Given the spefific heat of Cu is 0.385 J/g°C what will be the final temperature of the Cu
Solution
1
Calculate the heat absorbed: The heat absorbed by the copper block can be calculated using the formula q=m⋅c⋅ΔT q = m \cdot c \cdot \Delta T , where q q is the heat absorbed (in joules), m m is the mass (in grams), c c is the specific heat capacity (in J/g°C), and ΔT \Delta T is the change in temperature (in °C). Here, q=2.59 kJ=2590 J q = 2.59 \, \text{kJ} = 2590 \, \text{J} , m=22 g m = 22 \, \text{g} , and c=0.385 J/g°C c = 0.385 \, \text{J/g°C}
2
Rearrange the formula: To find the change in temperature ΔT \Delta T , we rearrange the formula to ΔT=qm⋅c \Delta T = \frac{q}{m \cdot c} . Substituting the values, we get: ΔT=2590 J22 g⋅0.385 J/g°C \Delta T = \frac{2590 \, \text{J}}{22 \, \text{g} \cdot 0.385 \, \text{J/g°C}}
3
Calculate \( \Delta T \): Performing the calculation gives: ΔT=259022⋅0.385≈303.5 °C \Delta T = \frac{2590}{22 \cdot 0.385} \approx 303.5 \, \text{°C}
4
Determine the final temperature: The final temperature Tf T_f can be found by adding the change in temperature to the initial temperature: Tf=Ti+ΔT=45 °C+303.5 °C≈348.5 °C T_f = T_i + \Delta T = 45 \, \text{°C} + 303.5 \, \text{°C} \approx 348.5 \, \text{°C}
Answer
The final temperature of the copper block is approximately 348.5 °C.
Key Concept
The specific heat capacity is a measure of how much heat energy is required to change the temperature of a substance.
Explanation
The final temperature is calculated by determining the heat absorbed and using the specific heat capacity to find the change in temperature.
A 22 g block of copper at 45 °C absorbs 2.50 kJ of heat. Given the spefific heat of Cu is 0.385 J/g°C what will be the final temperature of the Cu
Solution
1
Calculate the heat absorbed: The heat absorbed by the copper block can be calculated using the formula q=m⋅c⋅ΔT q = m \cdot c \cdot \Delta T , where q q is the heat absorbed (in joules), m m is the mass (in grams), c c is the specific heat capacity (in J/g°C), and ΔT \Delta T is the change in temperature (in °C). Here, q=2.50 kJ=2500 J q = 2.50 \, \text{kJ} = 2500 \, \text{J} , m=22 g m = 22 \, \text{g} , and c=0.385 J/g°C c = 0.385 \, \text{J/g°C}
2
Rearrange the formula to find the final temperature: We need to find the final temperature Tf T_f . First, we calculate ΔT \Delta T using the rearranged formula ΔT=qm⋅c \Delta T = \frac{q}{m \cdot c} . Substituting the values, we get ΔT=2500 J22 g⋅0.385 J/g°C \Delta T = \frac{2500 \, \text{J}}{22 \, \text{g} \cdot 0.385 \, \text{J/g°C}}
3
Calculate \( \Delta T \): Performing the calculation gives ΔT≈298.7 °C \Delta T \approx 298.7 \, \text{°C} . Now, we can find the final temperature Tf T_f using Tf=Ti+ΔT T_f = T_i + \Delta T , where Ti=45 °C T_i = 45 \, \text{°C} . Thus, Tf=45 °C+298.7 °C T_f = 45 \, \text{°C} + 298.7 \, \text{°C}
4
Final temperature: Therefore, the final temperature of the copper block is approximately Tf≈343.7 °C T_f \approx 343.7 \, \text{°C}
Answer
The final temperature of the copper block is approximately 343.7 °C.
Key Concept
The specific heat capacity is a measure of how much heat energy is required to raise the temperature of a substance.
Explanation
The final temperature is calculated by determining the heat absorbed and using it to find the change in temperature, which is then added to the initial temperature.
Solution
1
Calculate moles of HCl: To find the number of moles of HCl, use the formula: \\moles=Molarity×Volume (L) \text{moles} = \text{Molarity} \times \text{Volume (L)} \\. Here, the volume of HCl is 125 mL, which is 0.125 L, and the molarity is 1.750 M. Thus, \\moles of HCl=1.750 mol/L×0.125 L=0.21875 mol \text{moles of HCl} = 1.750 \, \text{mol/L} \times 0.125 \, \text{L} = 0.21875 \, \text{mol} \\
2
Calculate moles of NaOH: Similarly, for NaOH, the volume is 195 mL (0.195 L) and the molarity is 0.6667 M. Therefore, \\moles of NaOH=0.6667 mol/L×0.195 L=0.1300 mol \text{moles of NaOH} = 0.6667 \, \text{mol/L} \times 0.195 \, \text{L} = 0.1300 \, \text{mol} \\
3
Determine the limiting reactant: The reaction between HCl and NaOH is a 1:1 reaction. Since we have 0.21875 mol of HCl and 0.1300 mol of NaOH, NaOH is the limiting reactant
4
Calculate heat released: The heat released can be calculated using the heat of neutralization: \\q=moles of NaOH×ΔHn∘ q = \text{moles of NaOH} \times \Delta H^{\circ}_{\text{n}} \\. Substituting the values, \\q=0.1300 mol×(−56.2 kJ/mol)=−7.306 kJ q = 0.1300 \, \text{mol} \times (-56.2 \, \text{kJ/mol}) = -7.306 \, \text{kJ} \\
Answer
The heat released when 125 mL of 1.750 M HCl is mixed with 195 mL of 0.667 M NaOH is approximately -7.31 kJ.
Key Concept
The heat of neutralization is the heat change when an acid and a base react to form water and a salt.
Explanation
The calculation shows that the limiting reactant (NaOH) determines the amount of heat released in the reaction, which is negative indicating an exothermic process.
© 2023 AskSia.AI all rights reserved