Question

When 23.6g of calcium chloride were dissolved in 300mL of water in a calorimeter, the temperature of the water rose from 25°C to 38.7°C. What is the heat energy change in kcal for this process (The specific heat of H2O=1.00cal/g*°C).

Answer 1

Answer -

Given,

Volume of Water = 300 mL

Density = 1 g/mL

Mass of Calcium = 23.6 g

Initial Temperature = 25°C

Final Temperature = 38.7°C

Specific Heat of water = 1.00 cal/g.°C

We know that,

Density = Mass / Volume

Mass = Density * Volume

So, mass of 300 mL water = 1 g/mL * 300 mL = 300 g

Also,

Heat = m * c * T

Heat = 300 g * (1.00 cal/g.°C) * (38.7°C - 25°C)

Heat = 4110 cal

Now,

1 cal = 0.001 kcal

So, 4110 cal = 4110 * 0.001 kcal = 4.11 kcal [ANSWER]