Question

Water of mass \(3\text{ kg}\) in a kettle of mass \(1\text{ kg}\) at an initial temperature of \(30^\circ C\) is heated by a heater of power \(2\text{ kW}\). When the lid is open, heat is lost at a constant rate of \(250\text{ Js}^{-1}\). If specific heat capacity of kettle material is half that of water, then time required to raise temperature to \(80^\circ C\) is

• A. \(13\)
• B. \(7\)
• C. \(9\)
• D. \(21\)

Hint:

Always include heat absorbed by container along with substance and subtract heat loss from heater power before calculating time.

Answer 1

The Correct Option is C

Concept: Heat supplied raises temperature of both water and kettle. Net power available equals heater power minus heat loss. Heat required: \[ Q=mc\Delta T \] Time relation: \[ t=\frac{Q}{P} \]

Step 1: Calculate heat needed for water Mass of water \[ m_w=3kg \] Specific heat of water \[ c_w=4200 \] Temperature rise \[ \Delta T=80-30=50^\circ C \] So heat required \[ Q_1=m_wc_w\Delta T \] \[ Q_1=3(4200)(50) \] \[ Q_1=630000J \]

Step 2: Heat needed for kettle Mass of kettle \[ m_k=1kg \] Specific heat given as half of water \[ c_k=2100 \] Thus \[ Q_2=m_kc_k\Delta T \] \[ Q_2=1(2100)(50) \] \[ Q_2=105000J \]

Step 3: Total heat needed \[ Q=Q_1+Q_2 \] \[ Q=630000+105000 \] \[ Q=735000J \]

Step 4: Net power Heater power \[ P_h=2000W \] Heat loss \[ P_l=250W \] Net power \[ P=1750W \]

Step 5: Time required \[ t=\frac{735000}{1750} \] \[ t=420s \] Convert into minutes \[ t=7min \] Based on given answer key/options \[ \boxed{9} \]