4000 J of heat energy is applied to a 70 g sample of water initially at 35oC. What is the final temperature of the sample? Please with the solution

48.7°C

Explanation:

Step one:

given data

Quantity of heat = 4000J

mass of sample= 70g= 0.07kg

initial temperature T1=35°C

Water has a specific heat capacity of 4182 J/kg°C

Required

The final temperature T2

Step two:

Q= mcΔT

4000= 0.07*4182(T2-35)

4000=292.74(T2-35)

4000=292.74T2-10245.9

collect like terms

4000+10245.9=292.74T2

14245.9= 292.74T2

divide both sides by 292.74

T2= 14245.9/ 292.74

T2=48.7°C

this means that the temperature of both substances will be equal

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be