Final answer:
The short circuit current for a 120-V appliance cord with a 0.500-Ω resistance is 240 A. The energy dissipated over 0.0500 s is 1,440 J, leading to a temperature rise of approximately 859.8 °C in 2.00 g of surrounding materials with a specific heat capacity of 0.200 cal/g °C, which is likely damaging.
Step-by-step explanation:
Short Circuit Current Calculation
Given the values of voltage (V) and resistance (R), we can calculate the short circuit current (I) using Ohm’s law, I = V/R. For a 120-V appliance cord with a 0.500-Ω resistance, the short circuit current would be I = 120 V / 0.500 Ω = 240 A.
Temperature Rise Calculation
The energy (E) dissipated can be calculated using the power (P) and the time (t), where P = I² * R and E = P * t. For our calculated current, P = (240 A)² * 0.500 Ω = 28,800 W. The energy dissipated over 0.0500 s is E = 28,800 W * 0.0500 s = 1,440 J. To convert joules to calories, 1 J = 0.239005736 cal, so E = 1,440 J * 0.239005736 cal/J = 343.92 cal.
Using the specific heat capacity (c) and mass (m) of the materials, the temperature rise (ΔT) can be calculated with ΔT = E / (m * c). Substituting the values, ΔT = 343.92 cal / (2.00 g * 0.200 cal/g °C) = 859.8 °C.
This temperature rise is extremely high and is likely to be damaging to the surrounding materials of the appliance cord.