Physics

30 g of ice at 0°C is used to bring down the temperature of a certain mass of water at 70°C to 20°C. Find the mass of water.
[Specific heat capacity of water = 4.2 J g⁻¹ °C⁻¹ and specific latent heat of ice = 336 J g⁻¹]

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Given,

For ice

Mass of ice (m_i) = 30 g

Initial temperature (T_i) = 0°C

Final temperature (T_f) = 20°C

Specific latent heat of ice (L_i) = 336 J g⁻¹

For water

Initial temperature (T_i) = 70°C

Final temperature (T_f) = 20°C

Specific heat capacity of water (c_w) = 4.2 J g⁻¹ °C⁻¹

Let,

Mass of water = (m_w)

Now,

Heat lost by water = m_wc_w△t

= m_w × 4.2 × (70 - 20)

= m_w × 4.2 × 50 = 210m_w

Heat gained by ice = m_iL_i + m_ic_w△t

= 30 × 336 + 30 × 4.2 × (20-0)

= 10080 + 2520 = 12600 J

By principle of calorimetry,

Heat lost by water = Heat gained by ice

⟹ 210m_w = 12600

⟹ m_w = $\dfrac{12600}{210}$ = 6 g

So, the mass of required water is 6 g.

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