35 of copper pellets are removed from a 300 oven and immediately dropped into 120 of water at 20 in an insulated cup.What will the new water temperature be?
I'm stuck on this problem and could use some help. I used q pellets = m · Cp · ΔT and q water = m · Cp · ΔT to try and figure out the problem, but am not getting the right answer. Aren't these the equations i use and then solve for Tf. For some reason my numbers aren't working.
Any help would be great.
To solve this problem, you can apply the principle of conservation of energy, specifically the concept of heat transfer.
First, let's identify the information given:
- Initial temperature of copper pellets: 300 °C
- Mass of copper pellets: 35 g
- Initial temperature of water: 20 °C
- Volume of water: 120 mL (which can be assumed to be 120 g, as the density of water is approximately 1 g/mL)
- We need to find the final temperature of the water.
Now, let's approach this step by step:
Step 1: Calculate the heat transferred from the copper pellets to the water.
The equation you mentioned, q pellets = m · Cp · ΔT, is correct. However, you need to make sure you use the specific heat capacity (Cp) for each substance correctly.
The specific heat capacity of copper (Cp, copper) is approximately 0.38 J/g°C.
The specific heat capacity of water (Cp, water) is approximately 4.18 J/g°C.
Using the equation q pellets = m · Cp, copper · ΔT, where ΔT is the change in temperature, we have:
q pellets = (35 g) · (0.38 J/g°C) · (300 °C - Tf)
Step 2: Calculate the heat transferred to the water.
Similarly, using the equation q water = m · Cp, water · ΔT, we have:
q water = (120 g) · (4.18 J/g°C) · (Tf - 20 °C)
Step 3: Since energy is conserved, the heat transferred from the copper pellets equals the heat transferred to the water. Therefore, we can set q pellets = q water, and solve for Tf:
(35 g) · (0.38 J/g°C) · (300 °C - Tf) = (120 g) · (4.18 J/g°C) · (Tf - 20 °C)
Now, you can solve this equation for Tf, the final temperature of the water.
Note: Be careful with unit conversions and ensure you are using consistent units throughout the calculations.
If you're still having trouble getting the correct answer, double-check each step of your calculations and make sure you're using the correct values for Cp. Also, watch out for any rounding errors that may occur during the calculation process.
The sum of the heats gained is zero.
Heat gained by pellets+heatgained by water=0
masspellets(Cp)(Tf-300)+masswater(Cw)(Tf-20)=0
solve for Tf