Two 20.0-g ice cubes at –11.0 °c are placed into 205 g of water at 25.0 °c. assuming no energy is transferred to or from the surroundings, calculate the final temperature, tf, of the water after all the ice melts. heat capacity of h2o(l)= 75.3j/(mol*k) heat capacity of h2o(s)=37.7j/(mol*k) enthalpy of fusion of h2o=6.01kj/mol

The final temperature, T, of the water after all the ice melts = 6.9 °C

The law of conservation of energy can be applied to heat changes, i.e. the heat received / absorbed is the same as the heat released

Heat can be calculated using the formula:

Two 20-g ice cubes = 40-g / 18.0-g / mol = 2.22 mole of ice

205-g / 18.0g / mol = 11.39 mole water

Q1 = 2.22 x 37.7 J / (mol ° C) x 11

Q1 = 920.63 J

Q2 = mole x heat fusion

Q2 = 2.22 x 6.01kJ / mol

Q2 = 13.34 kJ = 13340 J

Q3 = 2.22 x 75.3J / (mol °C x (T-0)

Q3 = 167,166 T

Q4 = 11.39 x 75.3J / (mol ° C x (25-T)

Q4 = 857.67 (25-T)

Then:

920.63 J + 13340 + 167,166 T = 857.67 (25-T)

14260.63 + 167.166T = 21441.75 - 857.67T

857.67T + 167.166T = 21441.75 - 14260.63

1027.84 T = 7181.12

T= 6.98 C

the difference between temperature and heat

link

Specific heat

link

relationships among temperature, heat, and thermal energy.

link

When heat is added to a substance

link

Keywords: heat, temperature

може ли да зареждате повече информация, моля

the answers are

c and c