Question

Find the quantity of heat needed

to melt 100g of ice at -10 °C
into water at 10 °C. (39900 J)
(Note: Specific heat of ice is
2100 Jkg 'K', specific heat
of water is 4200 Jkg K',
Latent heat of fusion of ice is
336000 Jkg ').​

Answer 1

Approximately
`3.99* 10^(4)\; \rm J` (assuming that the melting point of ice is
`0\; \rm ^\circ C`.)

Step-by-step explanation:

Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.

`\begin{aligned}m&= 100\; \rm g * (1\; \rm kg)/(1000\; \rm g) \\ &= 0.100\; \rm kg\end{aligned}`

The energy required comes in three parts:

• Energy required to raise the temperature of that
  `0.100\; \rm kg` of ice from
  `(-10\; \rm ^\circ C)` to
  `0\; \rm ^\circ C` (the melting point of ice.)
• Energy required to turn
  `0.100\; \rm kg` of ice into water while temperature stayed constant.
• Energy required to raise the temperature of that newly-formed
  `0.100\; \rm kg` of water from
  `0\; \rm ^\circ C` to
  `10\;\ rm ^\circ C`.

The following equation gives the amount of energy
`Q` required to raise the temperature of a sample of mass
`m` and specific heat capacity
`c` by
`\Delta T`:

`Q = c \cdot m \cdot \Delta T`,

where

• 
  `c` is the specific heat capacity of the material,
• 
  `m` is the mass of the sample, and
• 
  `\Delta T` is the change in the temperature of this sample.

For the first part of energy input,
`c(\text{ice}) = 2100\; \rm J \cdot kg \cdot K^(-1)` whereas
`m = 0.100\; \rm kg`. Calculate the change in the temperature:

`\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (0\; \rm ^\circ C) - (-10\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}`.

Calculate the energy required to achieve that temperature change:

`\begin{aligned}Q_1 &= c(\text{ice}) \cdot m(\text{ice}) \cdot \Delta T\\ &= 2100\; \rm J \cdot kg \cdot K^(-1) \\ &\quad\quad * 0.100\; \rm kg * 10\; \rm K\\ &= 2.10* 10^(3)\; \rm J\end{aligned}`.

Similarly, for the third part of energy input,
`c(\text{water}) = 4200\; \rm J \cdot kg \cdot K^(-1)` whereas
`m = 0.100\; \rm kg`. Calculate the change in the temperature:

`\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (10\; \rm ^\circ C) - (0\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}`.

Calculate the energy required to achieve that temperature change:

`\begin{aligned}Q_3&= c(\text{water}) \cdot m(\text{water}) \cdot \Delta T\\ &= 4200\; \rm J \cdot kg \cdot K^(-1) \\ &\quad\quad * 0.100\; \rm kg * 10\; \rm K\\ &= 4.20* 10^(3)\; \rm J\end{aligned}`.

The second part of energy input requires a different equation. The energy
`Q` required to melt a sample of mass
`m` and latent heat of fusion
`L_\text{f}` is:

`Q = m \cdot L_\text{f}`.

Apply this equation to find the size of the second part of energy input:

`\begin{aligned}Q_2&= m \cdot L_\text{f}\\&= 0.100\; \rm kg * 3.36* 10^(5)\; \rm J\cdot kg^(-1) \\ &= 3.36* 10^(4)\; \rm J\end{aligned}`.

Find the sum of these three parts of energy:

`\begin{aligned}Q &= Q_1 + Q_2 + Q_3 = 3.99* 10^(4)\; \rm J\end{aligned}`.