Question

How many grams of ice at -14^oC are needed to cool 200g of water from 25^oC to 10^oC? (Take, specific heat of ice=0.5calg^-1oC^-1 and latent heat of ice =80calg^-1)

• A. 20g
• B. 27g
• C. 30g
• D. 31g

Answer 1

The correct option is D

31g

Step :1 Given data

Mass of water, $m_w=200g$

The initial temperature of ice, $T=-{14}^{o}C$

The initial temperature of the water, $T_1={25}^{o}C$

The final temperature of the water, $T_2={10}^{o}C$

Specific heat of ice, $c_i=0.5\frac{cal}{g{}^{o}C}$

Latent heat of ice, $L=80\frac{cal}{g}$

Step 2: Finding the amount of heat released

Heat energy $Q=mc∆T$

The heat released in cooling water from 25^oC to 10^oC is $Q_r=m_w{c}_w∆T$

$\Rightarrow Q_r=200g\times 1\frac{cal}{g{}^{o}C}\times \left({25}^{o}C-{10}^{o}C\right)$

$=200\times 15=3000cal$

Step :3 Finding the amount of heat absorbed

Heat absorbed by ice to convert into the water from -14^oC to 10^oC is

$Q_a=m_i{c}_i∆T+m_{i}L+m_i{c}_w∆T$

$Q_a=m_{i}g\times 0.5\frac{cal}{g{}^{o}C}\times \left(0^{o}C-\left(-{14}^{o}C\right)\right)+m_{i}g\times 80\frac{cal}{g}+m_{i}g\times 1\frac{cal}{g{}^{o}C}\times \left({10}^{o}C-0^{o}C\right)$

$=m_i\times 0.5\times 14+m_i\times 80+m_i\times 10$

$=7m_i+80m_i+10m_i=97m_{i}cal$

Step 4: Finding mass of ice required

According to the principle of calorimetry, the heat released by a body = heat absorbed by a body

Therefore, $Q_r=Q_a$ $\Rightarrow 3000=97m_i$

$\Rightarrow m_i=\frac{3000}{97}=30.9=31g$

Hence the mass of ice required is 31 g.

Thus D is the correct option.