Question

Two 20.0‑g ice cubes at −17.0 °C are placed into 225 g of water at 25.0 °C. Assuming no energy is transferred to or from the surroundings, calculate the final temperature of the water after all the ice melts. heat capacity of ?2?(?) 37.7 J/(mol⋅K) heat capacity of ?2?(?) 75.3 J/(mol⋅K) enthalpy of fusion of ?2? 6.01 kJ/mol

Answer 1

given

mass of ice cubes = 20 g

temperature of ice T1= -17

mass of water = 225 g

temperature of water T2 = 25

Let final temperature be T. Then

for ice to become water at temperature T first it should be converted to ice at 0 degrees and then water at 0 degrees and then water at T degrees.

The heat energy required to convert ice at -17 degrees to ice at 0 degrees H1 =

moles of ice* heat capacity of ice * ( final temperature - initial temperature)

=(20/18)*37.7*(0-(-17)) = 12818/18 =712 joules

Heat energy required to convert ice at 0 degrees to water at zero degrees H2 =mass of ice* enthalpy of fusion of ice

=(20/18)*6010 = 6677 j

heat energy to make ice into water at T degrees is H3

=moles of ice*heat capacity of water*(final temperature - initial temperature)

=(20/18)*75.3*(T-0)

Heat energy given by water H4 = moles of water*heat capacity of water*(initial temperature - final temperature )

= 225/18 * 75.3 *(25-T)

As no heat is from outside,

H4 = H1 + H2 +H3 +H4

23531- 941.25*T = 712 +6677 + 83.667T

T=16142/1024 = 15.76 degrees

HIT A LIKE THERE PLEASE!!!!!!!!

Answer 2

1. Given Data:

• Mass = 225 g

• Initial temperature = 25.0 °C

• Heat capacity of water,

• Moles of water = $\frac{225\text{g}}{18.015\text{g/mol}}=12.49\text{mol}$

• Mass = 2 × 20.0 g = 40.0 g

• Initial temperature = −17.0 °C

• Heat capacity of ice,

• Molar mass of ice (H₂O) = 18.015 g/mol → Moles of ice = $\frac{40.0\text{g}}{18.015\text{g/mol}}=2.22\text{mol}$

• Ice cubes:

• Water:

• Enthalpy of fusion of ice, $\mathrm{\Delta }{H}_{\text{fusion}}=6.01\text{kJ/mol}=6010\text{J/mol}$

2. Energy Required to Heat Ice to 0 °C:

   $$q_1=n_{\text{ice}}\times C_{\text{ice}}\times \mathrm{\Delta }T=2.22\text{mol}\times 37.7$$

4. Energy Required to Melt Ice:

   $$q_2=n_{\text{ice}}\times \mathrm{\Delta }{H}_{\text{fusion}}=2.22\text{mol}\times 6010\text{J/mol}=13,342\text{J}$$

6. Total Energy Absorbed by Ice ($q_{\text{absorbed}}$):

   $$q_{\text{absorbed}}=q_1+q_2=1422\text{J}+13,342\text{J}=14,764\text{J}$$

8. Energy Released by Water to Reach Final Temperature ($q_{\text{released}}$):

   $$q_{\text{released}}=n_{\text{water}}\times C_{\text{water}}\times (25.0\text{°C}-T_f)$$$$14,764\text{J}=12.49\text{mol}\times 75.3$$

10. Solve for Final Temperature ($T_f$):

    $$25.0-T_f=\frac{14,764}{12.49\times 75.3}$$$$25.0-T_f=15.7\text{°C}$$$$T_f=25.0\text{°C}-15.7\text{°C}=9.3\text{°C}$$

Answer:

The final temperature of the water after all the ice melts is 9.3 °C