A container contains water having mass 300g at 40° C. A piece of

SOLUTION

CONCEPT:

• Specific heat or specific heat capacity of a body is the amount of heat required for a unit mass of the body to raise the temperature by 1 degree Celsius.
  • It is denoted by “c” or “s”.
  • It is different for each substance.
  • Its unit is cal/g-°C in CGS and joule/g-K in SI unit.
  • 1 calorie = 4.186 Joule.
  • s = (Q /m Δ t) [s= specific heat of a substance, Q= Heat supplied to the object, m= mass of the substance, Δ t = change in temperature].

• Specific latent heat of fusion and vaporization:

• In the process of melting or vaporization, the temperature remains constant. The amount of heat needed to change the state of matter having mass m is-
• Q = m L [Q= heat required, m = mass of the body, L = latent heat]
• Latent heat of fusion: It is the heat required for 1 kg of substance to change it from solid to liquid state without change of temperature.
  • Latent heat of fusion of ice = 3.36 × 10⁵ J/kg
    • In CGS = 80 cal / gm.
• Latent heat of Vaporization: It is the heat required for 1 kg of substance to change it from liquid to gaseous state without change of temperature.
  • Latent heat of vaporization of water = 2.25 × 10⁶ J/kg.
    • In CGS = 536 cal / gm

EXPLANATION:

Heat is a form of energy it never destroys just transfer from one to another.

Given, the specific heat capacity of water = 4200 J/kg-K.

Specific latent heat of fusion of ice = 3.4 × 10⁵ J/kg.

Mass of water = 300g = 0.3 kg

Mass of Ice = 400g

The heat released as the water cools down from 40°C to 0°C.

Q = m s Δθ [Q = heat, m = mass of a body, s = specific heat of a body and Δθ = change in temperature]

Q = (0.3 kg) × (4200 J/kg-K) × (40 K)

Q = 50400 J.

The amount of ice melted by this much heat is given by

m = Q/L [m = mass of a body, Q = heat given, L = latent heat of body].

m = (50400 J) / (3.4 × 10⁵ J/kg.) ⇔ 0.1482 kg = 148g

Ice melted = 148g

Ice remaining = initial mass – ice melted

Ice left = 400g – 148g = 252g