Answer (A)
We will solve it by the equation
Q= m*C*(Tf-Ti)
Q= energy
m= mass of water= density*volume = 1g/mL*20 mL= 20 gram
c= specific heat of water = 4.186 j/gC
Tf= final temperature = 40 degree C
Ti= initial temperature = room temperature = 25 C
using the values
A) q= 20*4.186*(40-25) =1255.8 J (energy absorbed )
Answer (B)
all the data remain same but here Tf= 10 degrree C and Ti= 25 degrree C
using the values again
Q= 20*4.186*(10-25) = -1255.8 J (negative because energy is released )
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