A copper wire that has been heated to 520 degrees Celsius for pliability is plunnged into 350 mL of water at 18 degrees Celsius. The final temperature of the copper and water bath is 31 degrees Celsius. Find the mass of the copper wire
To find the mass of the copper wire, we can use the principle of conservation of energy.
The heat gained by the water and the copper wire is equal to the heat lost by the copper wire. We can use the formula:
m1 * c1 * ΔT1 = m2 * c2 * ΔT2
where:
m1 = mass of water
c1 = specific heat capacity of water
ΔT1 = change in temperature of water = final temperature - initial temperature of water
m2 = mass of copper wire
c2 = specific heat capacity of copper
ΔT2 = change in temperature of copper = final temperature - initial temperature of copper
Given values:
m1 = 350 mL = 350 g (since 1 mL of water is approximately equal to 1 g)
c1 = 4.18 J/g°C (specific heat capacity of water)
ΔT1 = 31°C - 18°C = 13°C
c2 = 0.39 J/g°C (specific heat capacity of copper)
ΔT2 = 31°C - 520°C = -489°C
Now we can rearrange the formula to solve for m2:
m2 = (m1 * c1 * ΔT1) / (c2 * ΔT2)
Substituting the given values:
m2 = (350 g * 4.18 J/g°C * 13°C) / (0.39 J/g°C * -489°C)
m2 = (6077 J) / (-190.71 J)
m2 ≈ -31.90 g
The negative mass doesn't make sense in this context, so we assume there might have been an error in the given data or calculations. Please double-check the values and ensure they are accurate.