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A sphere of alumininum of mass 0.047 kg ...

A sphere of alumininum of mass 0.047 kg placed for sufficient time in a vessel containing boling water, so that the sphere is at `100^(@)C`. It is then immediately transferred to 0.14 kg copper calorimeter containing 0.25 kg of water at `20^(@) C` . The temperature of water rises and attains a steady state at `23^(@)C` . calculate the specific heat capacity of aluminum. Specific heat capacity of copper = `0.386 xx 10^(3) J kg^(-1) K^(-1)`.
Specific heat capacity of water = `4.18 xx 10^(-3) J kg^(-1) K^(-1)`

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To solve the problem, we need to apply the principle of conservation of energy, which states that the heat lost by the aluminum sphere will be equal to the heat gained by the water and the copper calorimeter. ### Step-by-step Solution: 1. **Identify the Given Values:** - Mass of aluminum, \( m_{Al} = 0.047 \, \text{kg} \) - Initial temperature of aluminum, \( T_1 = 100 \, \text{°C} \) - Mass of copper, \( m_{Cu} = 0.14 \, \text{kg} \) ...
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A sphere of aluminium of 0.047 kg is placed for sufficient time in a vessel containing boiling water, so that the sphere is at 100^(@) C. It is then immediately transferred to 0.14 kg copper calorimeter containing 0.25 kg of water at 20^(@) C. The temperature of water rises and attains a steady state at 23^(@) Calculate the specific heat capacity of aluminium. (Give specific heat of copper =0.386xx10^(3)Jkg^(-1)K^(-1)) .

The specific heat capacity of water is :

The amount of heat energy required to convert 1 kg of ice at -10^@ C to water at 100^@ C is 7,77,000 J. Calculate the specific latent heat of ice. Specific heat capacity of ice = 2100 J kg^(-1) K^(-1) , specific heat capacity of water = 4200 J kg^(-1) K^(-1) .

A hot solid of mass 60 g at 100^(@)C is placed in 150 g of water at 20^(@)C . The final steady temperature recorded is 25^(@)C . Calculate the specific heat capacity of the solid. [Specific heat capacity of water = 4200 kg^(-1)""^(@)C^(-1) ]

A hot iron ball of mass 0.2 kg is added into 0.5 kg of water at 10^@ C. The resulting temperature is 30^@ C. Calculate the temperature of hot ball. Specific heat capacity of iron = 336 J kg^(-1) K^(-1) and specific heat capacity of water = 4.2 xx 10^3" "J" "kg^(-1) K^(-1) .

10 g of ice at 0°C absorbs 5460 J of heat energy to melt and change to water at 50^@ C. Calculate the specific latent heat of fusion of ice. Specific heat capacity of water is 4200 J kg^(-1) K^(-1) .

A 50 g block of metal is heated to 200^(@)C and then dropped into a beaker containing 0.5 kg of water initially at 20^(@)C. if the final equilibrium temperature of the mixed system is 22.4^(@)C , find the specific heat capacity of the metal. Given, specific heat capacity of water is 4182 J kg^(-1).^(@)C^(-1) .

A mass of 50 g of a certain metal at 150^@ C is immersed in 100 g of water at 11^@ C. The final temperature is 20^@ C. Calculate the specific heat capacity of the metal. Assume that the specific heat capacity of water is 4.2 g^(-1)K^(-1)

State in brief, the meaning of each of the following: The specific heat capacity of copper is 0.4 Jg^(1)""^@C^(-1)

NCERT ENGLISH-THERMAL PROPERTIES OF MATTER-EXERCISE
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