# What is the mass of the salt required to store heat?

## A tank of Glauber's salt is in a household. Its purpose is to store $2.8 \times {10}^{9} J$ of heat energy, and it can be heated from ${22}^{\circ} C$ to ${50}^{\circ} C$. The melting point of the salt is ${32}^{\circ} C$ and it has a ${L}_{\text{f}}$ (specific latent heat of fusion) of $241 \frac{k j}{k g}$. Its specific heat capacity in solid state is $1066 \frac{J}{k {j}^{\circ} C}$ and in liquid state is $1576 \frac{J}{k {j}^{\circ} C}$. This is all the information given. Please tell me how you get the values and the process to solving it. Thank you.

Jan 4, 2018

Solve for the mass using the heat equation

#### Explanation:

m = mass
s = specific heat
Q = total heat absorbed = $2.8 \times {10}^{9} J$
${T}_{i} , {T}_{m} , {T}_{f}$ are initial, melting and final temperature respectively.

1) Salt absorbs heat when its temperature rises, i.e. , $m s \Delta T$
2) While melting, it absorbs heat without temperature change i.e., $m {L}_{f}$.

Put these two together:
Total heat gained = Heat gained by salt from ${22}^{o} \to {32}^{o}$, heat gained while melting + heat gained from ${32}^{o} \to {50}^{o}$

$Q = m {s}_{\text{solid"(T_m-T_i)+mL_f+ms_"liquid}} \left({T}_{f} - {T}_{m}\right)$
$Q = m \left\{{s}_{\text{solid"(T_m-T_i)+L_f+s_"liquid}} \left({T}_{f} - {T}_{m}\right)\right\}$

Then solve for the mass.