Question

How much energy is needed to melt 15 grams of ice to water at $0^\circ C$ ?

Answer 1

To solve this question you must have the knowledge of specific heat and the latent heat. As in this question, phase change is taking place where solid ice melts to liquid therefore the term latent heat is applied.

Complete step by step answer:
It is given that the mass of ice is 15 grams
The temperature is $0^\circ C$.
The ice is melted by applying energy in the form of heat to change its state from solid to liquid. This process is called meting and the temperature at which the solid changes to liquid is the melting point temperature.
By doing so, the heat is being absorbed.
The relation between the heat absorbed and the temperature is given by a formula as shown below.
The formula is shown below.
$Q = m.c.\Delta T$
Where,
Q is the heat absorbed.
m is the mass
c is the specific heat
$\Delta T$ is the change in temperature.
The specific heat is defined as the amount one heat required per unit mass to raise the temperature by 1 degree Celsius.
As this phase is changing from solid to liquid, one new concept is introduced known as specific latent heat which is denoted by L.
The formula will be
$Q = m \times L$
The whole process is broken in two parts, first heat needed to melt the ice from $15^\circ C$ to $0^\circ C$ and second the heat needed to melt solid ice at $0^\circ C$to $0^\circ C$.
The specific heat of ice (C) is 2108 J/Kg$^\circ C$, the latent heat of ice (L) is $333 \times {10^3}J/kg$
$\Delta {Q_{Total}} = m.c.\Delta T + m.L$
$\Delta {Q_{total}} = \Delta {Q_{ice}} + \Delta {Q_{fusion}}$
And hence on solving,we have
$\Rightarrow \Delta {Q_{ice}} = (0.015kg) \times (2108J/Kg^\circ C) \times (0^\circ C - 15^\circ C)$
$\Rightarrow \Delta {Q_{fusion}} = (0.015kg) \times (333 \times {10^3})$
And again on doing the simplification,we have
$\Rightarrow \Delta {Q_{total}} = (0.015kg) \times (2108J/Kg^\circ C) \times (0^\circ C - 15^\circ C) + (0.015kg) \times (333 \times {10^3})$
$\Rightarrow \Delta {Q_{total}} = - 473.3J + 4995J$
And hence finally we have,
$\Rightarrow \Delta {Q_{total}} = 4520.7J$
Therefore, energy needed to melt 15 grams of ice to water at $0^\circ C$ is 4520.7J.

Note:
Make sure to convert the mass given in grams into kilograms as the specific heat and the latent heat is given in terms of J/kg. 1 grams is equal to $\dfrac{1}{{1000}}kg$.