Question

(a) You took two 49-g ice cubes from your −14 C kitchen freezer and placed them in 201 g of water in a thermally insulated container. If the water is initially at 24 C, what is the final temperature at thermal equilibrium? (b) What is the final temperature if only one ice cube is used? Hint: Watch out for the different possibilities!

Answer 1

a)

Q1 be the energy required to bring the water to 0C

Q1 = m1 c DT

Q1 = 4.2*201*24

Q1=20260.8 J

let Q2 nbe energy reqired to bring ice to 0C

Q2 = m2 C dT

Q2 = 2.03*2*14*49

Q2 =2785.16 J

energy required for both the cubes to melt = Q3 = m c DT

Q3 = 334*2*49

Q3 =32732 Joules

now since 32732+2785.16 > 20260.8

hence the amount of energy required for the ice cubes to melt is greater than that required for the water to come to 0C.

hence the final temperature is 0C.
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let the final temperature be T

accroingly from the principle of calorimetry

we have energy lost=energy gained

if there is only one ice cube, enthalpy of fusion of ice = 16366

enthalpy of reaching to 0C from -14C = 1392.58 J

so total energy required for the ice to melt = 16366 + 1392.58 =17758.6 J

from the principle of calorimtery

we get 4.2*201*(24-T) = 17758.6 + 4.2*49*(T-0)

Solving for T,

T= 2.38 C