Question

Calculate the mass of ice needed to cool 150g of water contained in a calorimeter of mass 50g at ${32}^{o}C$ such that the final temperature is $5^{o}C$.

(Take specific heat capacity of water $c_w=4.2J/g^{o}C$ and latent heat capacity of ice $L=330J/g$)

Answer 1

Step 1: Given data

Mass of calorimeter $m_C=50g$

Mass of water $m_w=150g$

Initial temperature $t_i={32}^{o}C$

Final temperature $t_f=5^{o}C$

Change in temperature $∆T={32}^{o}C-5^{o}C={27}^{o}C$

Specific heat capacity of water $c_w=4.2J/g^{o}C$

Latent heat capacity of ice $L=330J/g$

Step 2: Finding the mass of ice

According to the principle of the calorimeter:

$\mathrm{Heat}\mathrm{lost}=\mathrm{Heat}\mathrm{gained}$

$\mathrm{Heat}\mathrm{lost}\mathrm{by}\mathrm{water}={\mathrm{m}}_{\mathrm{w}}{\mathrm{c}}_{\mathrm{w}}∆\mathrm{T}$

$\mathrm{Heat}\mathrm{lost}\mathrm{by}\mathrm{water}=150\mathrm{g}\times 4.2\mathrm{J}/{\mathrm{g}}^{\mathrm{o}}\mathrm{C}\times {27}^{\mathrm{o}}\mathrm{C}$

$\mathrm{Heat}\mathrm{lost}\mathrm{by}\mathrm{water}=17010\mathrm{J}$

$\mathrm{Heat}\mathrm{gained}\mathrm{by}\mathrm{ice}={\mathrm{m}}_{\mathrm{i}}\mathrm{L}$

$\mathrm{Heat}\mathrm{gained}\mathrm{by}\mathrm{ice}={\mathrm{m}}_{\mathrm{i}}\times 330\mathrm{J}/\mathrm{g}$

From the principle of calorimeter:

${\mathrm{m}}_{\mathrm{i}}\times 330\mathrm{J}/\mathrm{g}=17010\mathrm{J}$

${\mathrm{m}}_{\mathrm{i}}=\frac{17010}{330}$

${\mathrm{m}}_{\mathrm{i}}=51.54\mathrm{g}$

Thus the mass of the ice needed to cool 150 g of water is 51.54 g.