Question

An electric heater of power 100W raises the temperature of 5kg of a liquid from 25⁰C to 31⁰C in 2 minutes. Calculate the specific heat capacity of the liquid​

Please solve it ASAP

Answer 1

Answer:

400J/Kg.℃

Explanation:

power P =100w

mass m=5kg

time T=2min=120 sec

Change in temperature ∆T =6℃

specific heat s= Q/m.∆T

Power P= i2R

Heat Q= i2Rt = P×t

therefore Q= 100×120/5×6=400J/kg.℃