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