Advertisements
Advertisements
प्रश्न
Calculate the mass of ice needed to cool 150 g of water contained in a calorimeter of mass 50 g at 32 °C such that the final temperature is 5 °C. Specific heat capacity of calorimeter = 0.4 J g-1 °C-1, Specific heat capacity of water = 4.2 J g-1°C-1, latent heat capacity of ice = 330 J g-1.
उत्तर
Heat lost by hot water = Heat gained by ice + Heat lost by calorimeter + Heat gained by melted ice
`("MC"Δ_t)_"water" + ("MC"Δ_t)_"calorimeter" = "MLL"_"F" + ("MC"Δ_t)_"Melted ice"`
∴ 150 × 4.2 (32 - 5) + 50 × 0.4 × (32 - 5) = M × 330 + M × 4.2 × (5 - 0)
∴ (150 × 4.2 × 27) + (50 × 0.4 × 27) = 330 M + 21 M
17010 + 540 = 351 M
∴ M = `17550/351`
∴ M = 50
Mass of ice required = 50g.
APPEARS IN
संबंधित प्रश्न
Water is used in hot water bottles for fomentation. Give a reason.
State the impact of global warming on life on the earth.
Define the term 'specific heat capacity' and state its unit.
Specific heat capacity of a substance A is 3.8 J g-1 K-1 and of substance B is 0.4 J g-1 k-1. Which substance is a good conductor of heat? How did you arrive at your conclusion?
Write two advantages of high specific heat capacity of water.
A monoatomic gas of pressure 'P' having volume 'V' expands isothermally to a volume '2V' and then adiabatically to a volume '16V'. The final pressure of the gas is ______.
(ratio of specific heats = `5/3`)
At same temperature and pressure of an ideal gas, ____________.
Find the odd one out:
Observe the following diagram and answer the questions given below:
Specific heat capacity of metals
- Which element has maximum specific heat capacity? Justify.
- Which element has minimum specific heat capacity? Justify.
- Define specific heat of object.
