Question

6

Answer 1

Answer:

Detail is given below.

Explanation:

Given data:

Mass of water = 50 g

Initial temperature = 0°C

Final temperature = 100°C

Specific heat capacity of water = 1 cal/g.°C

Calories absorbed = ?

Solution:

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 - 0°C =  100°C

Now we will put the values in formula.

Q = 50 g×1 cal/g.°C ×100°C

Q = 5000 cal

Hence, it is proved that 5000 calories required to increase the temperature of 50 g of water from 0°C to 100°C.