Question

Calculate the temperature change when 1000J of heat is supplied to 100g of water.
Please explain

Answer 1

Answer:

It is sensible heat- the amount of heat absorbed by 1 kg of water when heated at a constant pressure from freezing point 0 degree Celsius to the temperature of formation of steam i.e. saturation temperature

So it is given as - mass× specific heat × rise in temperature

i.e. 4.2 × T

4.2 × (100–0)

So it is 420kj

If you ask how much quantity of heat is required to convert 1 kg of ice into vapour then you have to add latent heat of fusion that is 336 kj/kg and latent heat of vaporization 2257 kj/kg (these two process occur at constant temperature so need to add rise in tempeature)

So it will be

Q= 1×336 + 1× 4.18 ×100 + 1× 2257

Q = 3011 kj

Or 3.1 Mj

Hope you got this!!!!!!

Answer 2

Explanation:

Given

heat supplied (Q) = 1000 J

mass(m) = 100 g = 0.1 kg

specific heat capacity of water(s) = 4200 J/kg°C

change in Temperature (dt) = ?

We know we have the relation

Q = m * s * dt

Or, 1000 = 0.1 * 4200 * dt

Or, 1000 = 420 dt

Or, dt = 1000/ 420

Therefore dt = 2.38°

Hope it will help :)