How many calories is required to change the temperature of 2.18g of water from 15.3°C to 69.5°C. The specific heat of liquid water is 1.00cal/g°C.

Given:

Mass of water, m = 2.18 g

Initial temperature of water, T1 = 15.3 C

Final temperature, T2 = 69.5 C

Specific heat, c = 1.00 cal/gC

To determine:

Heat required to raise the water temperature

Explanation:

The amount of heat, Q, required to raise the temperature of water from T1 to T2 is:

Q = mcΔT = mc (T2-T1)

Q = 2.18 * 1 * (69.5-15.3) = 118.16 cal

Ans: Heat required is 118.16 cal

The number of calories that are required to change the temperature of 2.18 g of water from 15.3 c to 69.5 c is 118.16 cal

calculation

Heat in calories = MCΔ T where, M (mass) = 2.18 g C (specific heat capacity) = 1.00 cal/g/c ΔT (change in temperature) = 69.5 - 15.3 = 54.2 c

heat is therefore = 2.18 g x 1.00 cal/g/c x 54.2 c=118.16 cal