230 cm3 of hot tea at 96 °c are poured into a very thin paper cup with 100 g of crushed ice at 0 °c. calculate the final temperature of the "ice tea.

Final temperature of the ice tea, x = 42.73°C

Explanation:

230 cm^3 = 230 g of liquid ( taking the density of tea as the same the density of water)

In cooling from 96°C to 0°

C (before freezing) the available energy is

96 × 230 = 22080 cal

The latent heat (the heat required to melt the crushed ice per gram) of fusion for water = 79.7 cal/g

Melting 100 g of crushed ice at 0°C will absorb

100 × 79.7 = 7970 cal

The retain heat of the iced Tea = (22080 - 7970) cal of heat raising the mixture temperature as follows

(22080 - 7970) = 14110 cal

The mass of the final solution 230 + 100 = 330 g solution

Final temperature = 14110 / 330 = 42.76°C

Or where specific heat of water = 4.186J/Kg°C

and the latent heat of the ice = 334J/Kg

We have

230 × (specific heat capacity of water) × (96 - x) =100× (specific heat capacity of water) ×(0-x) + (latent heat) ×100 =

230 × 4.186 × (96 - x) =100× 4.186 ×(x-0) + 100×334

1381.38×x = 59026.88

Final temperature of the ice tea, x = 42.73°C

"record the time to complete a chemical reaction" would most likely be part of a lab procedure.