If 10 calories of energy are added to 2 grams of ice at -30° c, calculate the final temperature of the ice. (notice that the specific heat of ice is different from that of water.) t. f =
-20° c
-30° c
30° c
40° c

-20°C

Explanation:

ΔQ = m c Δt

here ΔQ is heat supplied which is 10 calories , m is mass which is 2 g , c is specific heat capacity of ice which is 0.5 cal/(g C°), Δt is change in temperature which here is to be calculated.

re arranging the equation  we get

Δt = ΔQ/(mc)

   = 10/ ( 2 × 0.5)

 =  10/1

  = 10 C°

final temperature = -30 +10 = -20°C

hence option is A

a and c

c is 100% but i am confused in a

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr