IF A HEATER WITH POWER 1000W GIVES 3.6*10^6 joules OF HEAT ENERGY IN ONE HOUR.WHAT WILL BE THE TEMPERATURE RAISED IN THE 50KG OF WATER TEMPERATURE 30 DEGREE CELCIUS BY THAT HEAT.

 

Asked by anish.kunwar.123 | 17th Jul, 2018, 09:11: PM

Expert Answer:

Heat required to raise the temperature of water is Q= 3.6×106 J.
Mass of water = 50 kg 
 
We know that Q = m×c×Δt ... (1) 
Rise in temperature Δt= tf - ti
A the temperature of water is 30°C,
Δt= tf - 30
Specific heat of water is 4200J/kg/°C.
 
From (1),
 
rightwards double arrow 3.6 space cross times 10 to the power of 6 space equals 50 space cross times 4200 cross times open parentheses t subscript f space minus 30 space close parentheses
rightwards double arrow space t subscript f space equals space fraction numerator 3.6 space cross times 10 to the power of 6 over denominator 50 space cross times 4200 end fraction space plus 30
rightwards double arrow space t subscript f space equals space 47.14 to the power of omicron C
temperature of water raised due to this amount of energy is ( tf)= 47.14°C
 
→ Δt = 47.14°C - 30°C = 17.14°C
 
Thus, the rise in temperature is 17.14°C

Answered by Shiwani Sawant | 18th Jul, 2018, 12:21: PM