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Question
A solid of mass 50 g at 150°C is placed in 100 g of water at 11°C when the final temperature recorded is 20°C. Find the specific heat capacity of the solid. (specific heat capacity of water = 4.2 J/g °C)
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Solution
Mass of the solid, ms = 50 g
Initial temperature of the solid, ts = 150°C
Mass of water, mw = 100 g
Temperature of water, tw = 11°C
Final temperature of the mixture, t = 20°C
According to the principle of calorimetry
Heat gained by water = Heat lost by the solid
`:.m_wc_w(t - t_w) = m_sc_s(t_s - t)`
∴ 100 × 4.2 × (20 - 11) = 50 × cs × (150 - 20)
3780 = 6500 cs
`c_s = 3780/6500`
`∴ c_s = 0.58` J/g °C
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| 1. | Specific heat capacity of water | a. | 0°C |
| 2. | Latent heat of fusion of ice | b. | 2260 J/g |
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