Physics · Heat, temperature, thermal expansion, specific heat capacity, calorimetry, change of state, latent heat

It takes 15 minutes for an electrical kettle to heat a certain quantity of water

It takes 15 minutes for an electrical kettle to heat a certain quantity of water in grams from \( 0^{\circ} \mathrm{C} \) to boiling point \( 100^{\circ} \mathrm{C} . \) If it takes 81 minutes to boil water into steam, the latent heat of steam is :

  • A. 80 cal \( l \),
  • B. 540 cal \( / g \)
  • C. 100 cal \( / g \)
  • D. 336 cal \( / g \)

Step-by-step solution

The kettle supplies constant power P. For heating: Q_heat = m * c * ΔT = m * 1 * 100 = 100m cal, time = 15 min, so P = (100m)/15. For boiling: Q_vap = m * L, time = 81 min, so P = (m * L)/81. Equating: (100m)/15 = (m * L)/81 → L = (100 * 81)/15 = 540 cal/g.
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