Two 69 g ice cubes are dropped into 230 g of water in a thermally insulated cont

Question

Two 69 g ice cubes are dropped into 230 g of water in a thermally insulated container. If the water is initially at 25 degree C, and the ice comes directly from a freezer at -15 degree C, what is the final temperature at thermal equilibrium? (Neglect the heat capacity of the glass.) What is the final temperature if only one ice cube is used? You'll probably need to guess at the final state (is any ice left?) and then check the guess with a calculation. The total energy remains constant. So, the energy decrease of the liquid must match the energy increase of the ice (and the consequent liquid if all the ice melts).

Explanation / Answer

Since you have got the first one right, I will solve the second one only.

b)Let the final temperature be T.

Heat gained by ice=mc1T+mL+mc2T

c1=specific heat capacity of ice.

c2=specific heat capacity of water.

=0.069*2108*15+0.069*334000+m*4186*T

Heat lost by water=m'c2T=0.23*4186*(25-T)

Heat lost by water=Heat gained by ice

0.23*4186*(25-T)=0.069*2108*15+0.069*334000+0.069*4186*T

Solving for T,

T=-0.93C

Since T<0 which is not possible for the heat balance we did, the final temperature is 0 C

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.