Question

A

Answer 1

Answer:

$\Delta T=3.615^{\circ}C$ is the drop in the water temperature.

Explanation:

Given:

• mass of ice, $m_i=14.7\ g=0.0147\ kg$

• mass of water, $m_w=324\ g=0.324\ kg$

Assuming the initial temperature of the ice to be 0° C.

Apply the conservation of energy:

• Heat absorbed by the ice for melting is equal to the heat lost from water to melt ice.

Now from the heat equation:

$Q_i=Q_w$

$m_i.L=m_w.c_w.\Delta T$ ......................(1)

where:

$L=$ latent heat of fusion of ice $=333.55\ J.g^{-1}$

$c_w=$ specific heat of water $=4.186\ J.g^{-1}.^{\circ}C^{-1}$

$\Delta T=$ change in temperature

Putting values in eq. (1):

$14.7 \times 333.55=324\times 4.186\times \Delta T$

$\Delta T=3.615^{\circ}C$ is the drop in the water temperature.