Question

We drop a cube of ice into a glass of water. The mass of the cube of ice is40.0 g and its initial temperature is −10.2∘C−10.2∘C. The mass of the water is 251 g and its initial temperature is 19.7∘C19.7∘C. What is the final temperature of the water after all of the ice has melted?

• ??????=4186?/??⋅∘?cwater=4186J/kg⋅∘C ????=2090?/??⋅∘?cice=2090J/kg⋅∘C ?????=128?/??⋅∘?clead=128J/kg⋅∘C
• ??????=380?/??⋅∘?cbrass=380J/kg⋅∘C ??????−???=3.33×105?/??

Answer 1

Amount of heat required to get converted ice from - 10.2⁰ C to 0⁰ C water = m x c x dT + mL;

= 0.04 x 2090 x (10.2) + 0.04 x 333000

= 14172.72 J.

Amount of heat released to get converted 19.7⁰ C to 0⁰ C water = M x C x dT;

= 0.251 x 4186 x (19.7);

= 20698.514 J.

Balance heat Q = 20698.514 - 14172.72;

= 6525.8 J

Now the final Temperature is T;

Q = (M + m) x C x dT;

6525.8 = (0.251 + 0.04) x 4186 x (T - 0);

6525.8 = 0.291 x 4186 x (T - 0);

T = 5.357⁰ C

Answer 2

Answer:

Final temperature of the water after the ice melts: ~10.6°C

Key Steps:

1. Heat absorbed by ice:

• Warms from -10.2°C to 0°C: $Q_1=m_{\text{ice}}\cdot c_{\text{ice}}\cdot \mathrm{\Delta }T=0.040\cdot 2090\cdot 10.2=852.72\text{J}$.

• Melts at 0°C: $Q_2=m_{\text{ice}}\cdot L_f=0.040\cdot 3.33\times 10^5=13,320\text{J}$.

• 
  

2. Heat lost by water:

• Cools to final temperature $T_f$: $Q_3=m_{\text{water}}\cdot c_{\text{water}}\cdot (19.7-T_f)$.

3. Energy balance ($Q_1+Q_2=Q_3$):

   $$852.72+13,320=0.251\cdot 4186\cdot (19.7-T_f)$$

   Solve for $T_f$:

   $$T_f\approx 10.6\mathrm{°}C$$

Assumptions: No heat loss to surroundings.