280 g of ice at 0°C is mixed with 600 g of water at 50°C present in a metal container of mass 300 g.
If the final temperature of the mixture is 10°C. Assuming no loss of heat to the atmosphere, calculate the specific heat capacity of the metal container.
(Specific latent heat of ice is 336 J g-1,
Specific heat capacity of water is 4.2 J g-1 K-1)
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Given,
- Mass of ice (m1) = 280 g at 0 °C
- Mass of water (m2) = 600 g at 50 °C
- Mass of the metal container (m3) = 300 g at 50 °C
- Final temperature of the mixture = 10 °C
- Specific latent heat of ice (L) = 336 J g-1
- Specific heat capacity of water (cw) = 4.2 J g-1 K-1
Let the specific heat capacity of the metal container be c.
Heat energy gained by the ice :
Heat used in melting the ice = m1 x L = 280 x 336 = 94080 J
Heat used in raising the temperature of the melted water from 0 °C to 10 °C
= m1 x cw x ΔT
= 280 x 4.2 x 10
= 11760 J
Total heat energy gained = 94080 + 11760 = 105840 J
Heat energy lost by the water and the container :
Heat lost by water = m2 x cw x ΔT
= 600 x 4.2 x (50 - 10)
= 600 x 4.2 x 40
= 100800 J
Heat lost by the container = m3 x c x ΔT
= 300 x c x 40
= 12000 c
By the principle of calorimetry,
Heat energy gained = Heat energy lost
$$\begin{aligned}105840 = 100800 + 12000\ \text c \\[1em] \Rightarrow 12000\ \text c = 105840 - 100800 \\[1em] \Rightarrow 12000\ \text c = 5040 \\[1em] \Rightarrow \text c = \frac{5040}{12000} = 0.42\end{aligned}$$
Hence, the specific heat capacity of the metal container is 0.42 J g-1 K-1.