The temperature of 170g of water at $50^{ο} C$ is lowered to $5^{ο} C$ by adding certain amount of ice to it. Find the mass of the ice added.
(Given: Specific heat capacity of water $= 4200 J K g^{- 1} {}^{ο}C^{- 1}$ and specific latent heat of ice $= 33600 J K g^{- 1}$ )

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Solution

Step 1: Given data
Mass of water $m_{w} = 170 g = 0.17 k g$
Temperature $T_{1} = 50^{ο} C$
Temperature $T_{2} = 0^{ο} C$
Temperature $T = 5^{ο} C$
Specific heat capacity of water $c_{w} = 4200 J K g^{- 1} {}^{ο}C^{- 1}$
Specific latent heat of ice $c_{i} = 336000 J K g^{- 1}$
Step 2: Finding the mass of the ice
According to the principle of Calorimetry: Heat lost by the hotter body= heat gained by the colder body
Let m_i be the mass of ice.
On applying the principle of Calorimetry:
$m_{w} \times c_{w} \times (T_{1} - T) = m_{i} \times c_{i} \times (T - T_{2})$
$0.17 g \times 4200 J K g^{- 1} {}^{ο}C^{- 1} \times (50^{ο} C - 5^{ο} C) = m_{i} \times 336000 J K g^{- 1} + m_{i} (4200 J K g^{- 1} {}^{ο}C^{- 1} \times 5^{ο} C)$
$32130 = 357000 m_{i}$
Then the mass of the ice added is:
$m_{i} = \frac{32130}{357000} = 0.09 k g$
Thus the mass of the ice added is 0.09 kg.

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Q(c). The temperature of 170g of water at 50°C is lowered to 5°C by adding certain amount of ice to it. Find the mass of ice added.
Given: Specific heat capacity of water = 4200 J kg^–1°C^–1 and specific latent heat of ice =336000 J kg^–1

By adding some amount of ice, the temperature of 500g of water at $50^{\circ} C$ , is lowered to $0^{\circ} C$ . If the specific heat capacity of water is 4.2 J g⁻¹ K⁻¹and specific specific latent heat of ice is 336 J g⁻¹, then find the mass of ice added to it.