Question

How much ice at $- 10 ° C$ must be dropped into a cup containing $300 g$ of water at $20 ° C$ in order for the temperature of the water to be reduced to $10 ° C$ ? The cup itself has a mass of $150 g$ and is made out of aluminium. Assume that no energy is lost to the surroundings.

$($ Specific heat capacity of ice $= 2200 J {kg}^{- 1} K^{- 1};$ specific latent heat of fusion of ice $= 334 J {kg}^{- 1} K^{- 1};$ specific heat capacity of water $= 4200 J {kg}^{- 1} K^{- 1} .)$

Accepted Answer

Given,

Mass of water, $m_{1} = 300 g = 0.3 kg$

Specific heat of water, $150 g$

Specific heat of aluminium, $s_{2} = 900 J {kg}^{- 1} K^{- 1}$

Mass of cup, $m_{2} = 150 g = 0.15 kg$

$∆ T = 10 ° C = 10 K$

Let $m_{3}$ be the amount of ice needed.

Let $Q$ be the energy transferred from the water and aluminium container.

$Q = m_{1} s_{1} ∆ T + m_{2} s_{2} ∆ T$

Substituting all the values,

$Q = 0.3 \times 4200 \times 10 + 0.15 \times 900 \times 10$

$\Rightarrow Q = 13950 J$

As there is no energy lost to the surroundings,

Therefore, this thermal energy will be taken up by the ice to $:$

$(i)$ Increase its temperature from $- 10 ° C$ to $0 ° C :$ the thermal energy required is $m_{3} \times 2200 \times 10 J$

$(ii)$ Melt the ice cube into water at $0 ° C :$ the thermal energy required is $m_{3} \times 334 \times 10^{3} J$

$(iii)$ Increase the temperature of the former ice cube from $0 ° C$ to the final temperature $10 ° C$ $:$ the thermal energy required is $m_{3} \times 4200 \times 10 J$ .

Thus $:$

From the principle of calorimetry,

$13950 = m_{3} \times 2200 \times 10 + m_{3} \times 334 \times 1000 + m_{3} \times 4200 \times 10$

$\Rightarrow 13950 = m_{3} \times (22000 + 334000 + 42000)$

$\Rightarrow m_{3} = \frac{13950}{398000} = 0.035 kg = 35 g$

So $35 g$ of ice is required.

K A Tsokos Solutions for Chapter: Thermal Physics, Exercise 1: Test yourself

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