Hi tick the correct option in each case

The curve shows it is almost reaching the surrounding temperature 20 °C and change in temperature keep on decreasing and it is very less at   4  hour.  Hence, temperature will be very near to 20 °C

 

Energy loss  ΔE for change in temperature ΔT is given by  ΔE = m×C_p×ΔT ,  

where m is mass of water and C_p is specific heat of water

if  ΔT = 1°C   ,  then Energy loss per °C,    m×Cp = 8×10⁵ J /°C .................(1)

 

Rate of change of energy loss with repect to time,  (ΔE/Δt) = m×C_p×( ΔT/Δt) = 16×10⁶ J/hour  ................(2)

 

from eqn.(1) and (2),  Rate of temperature falling,  ( ΔT/Δt) = 16×10⁶ / ( 8×10⁵)  = 20 °C/hour