Question

$0.1kg$ of ice at $0^\circ C$ is heated up to water at $30^\circ C$. Calculate the heat absorbed in this process. (take ${c_w} = 4186{\text{J/kg}}^\circ {\text{C}}$ and ${L_i} = 334000{\text{J/kg}}$)

Answer 1

Since the state was at ice, the latent heat required to melt the ice to water at the same temperature must be considered. The total heat absorbed is the sum of the latent heat plus the heat required to raise the temperature of water from one temperature to another.

Formula used: In this solution we will be using the following formulae;
${H_l} = mL$is the heat required to melt a substance from its solid state at a particular temperature to its liquid state at the same temperature, $m$ is the mass of the substance, and $L$ is called the latent heat of fusion.
$H = mc\Delta T$ where $H$ is the heat required to raise the temperature by an amount $\Delta T$, where $m$ is the mass of the substance, $c$ is the specific heat (also called specific heat capacity) of the substance.

Complete Step-by-Step Solution:
To calculate the total heat absorbed by the substance, we must calculate the heat required to melt the ice to water, then the heat required to raise the temperature to $30^\circ C$.
The heat required to melt substance from solid to liquid is given as
${H_l} = mL$ where $m$ is the mass of the substance, and $L$ is called the latent heat of fusion.
Hence, for ice as given
${H_{li}} = {m_i}{L_i} = 0.1 \times 3.34 \times {10^5} = 3.34 \times {10^5}{\text{J}}$
For heat absorbed by a substance until the temperature changes by an amount can be given as
$H = mc\Delta T$ where $H$ is the heat required to raise the temperature by an amount $\Delta T$, where $m$ is the mass of the substance, $c$ is the specific heat (also called specific heat capacity) of the substance.
Hence, for water, we have
$H = {m_w}{c_w}\Delta T = 0.1 \times 4186 \times \left( {30 - 0} \right)$
By computation, we get,
$H = 12558{\text{J}}$
Hence, the total heat absorbed is given by
${H_T} = H + {H_l}$
$\Rightarrow {H_T} = 12558 + 334000 = 346558{\text{J}}$

Note: For clarity, observe that we do not have to convert the temperature to Kelvin. This is for two reasons, one is the unit of the specific heat given to us is in ${\text{J/kg}}^\circ {\text{C}}$ and hence we can use the temperature as degree Celsius since any conversion factor will cancel. Secondly the unit ${\text{J/kg}}^\circ {\text{C}}$ is equivalent to ${\text{J/kg - K}}$, and this is because in the heat equation, the difference is what matters and not the temperature itself.