Question

Calculate the mass of ice required to lower the temperature of 300g of water at 40°C to water at 0°C.
(Specific latent heat of we = 336 J/g, Specific heat capacity of water = 4.2 J/g°C)​

Answer 1

Heat extracted by ice at 0⁰C to melt = Heat given by water to reach 0⁰C from 40⁰C.

$\boxed{\sf {ML=mcθ}}\orange\bigstar$

Where,  

• θ = change in temperature of water
• L = Specific Latent heat of ice
• c = Specific heat capacity of water
• M = Mass of ice required
• m = mass of water

$\sf{M×336=300×4.2×(40−0)}$

$\sf{M= \frac{300×4.2×40}{366}}$

$\: \: \: \: \: \: \: \: \: \boxed{ \frak {M=150g}} \pink\bigstar$

∴ Ice required to lower the temperature of water:  150g

Thank you!

@itzshivani

Answer 2

Mass of water = 300 gm

Specific heat of water = 4.2 J/g°C

Specific heat of ice = 336 J/g

Fall in temperature, θF = 40 - 0 = 40°C

Heat gained by ice = Heat lost by water

∴ Mass of ice x Specific heat of ice = Mass of water x Specific heat of water x Fall in temperature

Mass of ice = Mass of water x water x specific heat of water x θF/Specific heat of ice

=300x4.2x40 / 336

=150gm

Explanation:

Hope it will help you..