Question

Ice at 0.0 degrees Celsius is combined with 50.0g of water at 75.0 degrees Celsius. Calculate the grams of ice present initially if the entire mixture comes to a final temperature of 25.00 degrees Celsius after the ice melts. Specific heat of water is 4.18 J/g degrees Celsius. SHOW WORK!

Answer 1

m-Ice = 23.85g

Step-by-step explanation:

The heat absorbed for the cold water (Heat of the increase of water + Heat of ice melting) is equal to the heat released for the water at 75°C. The equation is:

m-Ice-*C*ΔT + Hf*-m-Ice = m-water*C*ΔT

Where m-Ice- is our incognite

C is specific heat of water = 4.18J/g°C

ΔT is change in temperature = 25.0°C - 0.0°C = 25.0°C

Hf is enthalpy of fusion of water = 333.6J/g

m-water = 50.0g

ΔT is change in temperature = 75.0°C - 25.0°C = 50.0°C

m-Ice-*4.18J/g°C*25°C + 333.6J/g*-m-Ice = 50.0g*4.18J/g°C*50.0°C

104.5 m-Ice- + 333.6 m-Ice- = 10450

438.1 m-Ice = 10450

m-Ice = 23.85g