A 500 W electric immersion heater is placed into 1.0 kg of water at 20 °C. Assuming negligible heat losses and specific heat capacity of water c = 4200 J/(kg*°C), how long will it take to heat the water to 70 °C?
Answer and explanation
Correct answer: D. 420 seconds (7 minutes)
- Step 1: Calculate thermal energy required: Q = m c Delta_T = 1.0 kg 4200 J/(kg°C) * (70 - 20) °C.
- Step 2: Compute Q: 4200 * 50 = 210,000 J.
- Step 3: Relate electrical energy to power and time: E = P * t => t = Q / P.
- Step 4: Compute time: t = 210,000 J / 500 W = 420 seconds = 7.0 minutes.
Answer
420 seconds (7 minutes).
Why not the others: '210 seconds (3.5 minutes)' divides by double the power or halves the temperature delta; '840 seconds (14 minutes)' doubles the required thermal energy; '350 seconds (5.8 minutes)' uses specific heat of 3500 J/(kg*°C).
Practise more Physical and Chemical Principles.
Free questions with an explanation for every one, and full timed Jubilee Institute Admission Test mock tests.
More Physical and Chemical Principles questions
- A uniform rod of length 2.0 m and mass 6.0 kg is pivoted at one end. A vertical upward force F is applied at the opposite end to keep the ro…
- An aqueous solution of hydrochloric acid has a pH of 2.0. If 10 mL of this solution is diluted with pure distilled water to a final volume o…
- A car traveling at a constant speed of 20 m/s accelerates uniformly at 2.5 m/s^2 over a straight distance of 100 m. What is the final veloci…
- Which of the following electronic configurations represents a transition metal atom in its ground state according to Hund's rule and the Auf…
- Two point charges of +4.0 microcoulombs and -1.0 microcoulombs are positioned 0.30 m apart along a line in a vacuum. At what point along the…
- When 5.4 g of solid aluminum (molar mass = 27 g/mol) reacts completely with excess iron(III) oxide via the thermite reaction: 2Al + Fe2O3 ->…