Question

2. 100 g of liquid water is at 10°С. Find the number of calories it takes to heat this up this liquid water to

100°C using this information.
100°C
10°C

Answer 1

8.99 Kcal

Step-by-step explanation:

Given data:

Mass of water = 100 g

Initial temperature = 10°C

Final temperature = 100°C

Calories take = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Specific heat capacity = 4.18 J/g.°C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 100°C - 10°C

ΔT = 90 °C

Q = 100 g × 4.18 J/g.°C × 90 °C

Q = 37620 J

Joule to calories:

37620 J × 1 kcal / 4184 J

8.99 Kcal