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420 J of energy supplied to 10g of water...

`420 J` of energy supplied to `10g` of water will rises its temperature by

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To solve the problem of how much the temperature of 10 g of water will rise when 420 J of energy is supplied, we can use the formula related to heat transfer: ### Step-by-Step Solution: 1. **Identify the Formula**: The heat supplied (ΔQ) is related to the mass (m), specific heat capacity (s), and the change in temperature (ΔT) by the formula: \[ \Delta Q = m \cdot s \cdot \Delta T ...
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When 60 calories of heat are supplied to 15 g of water, the rise in temperature is

50 g of copper is heated to increase its temperature by 10^@C . If the same quantity of heat is given to 10g of water, the rise in its temperature is (specific heat of copper =420J//kg^(@)//C )

50 gm of copper is heated to increase its temperature by 10^(@)C. If the same quantity of heat is given to 10 gm of water, the rise in its temperature is (Specific heat of copper =420 Joule-kg^(-1)@C^(-1)

A plate of area 10 cm^2 is to be electroplated with copper (density 9000 kg m^(-3) to a thickness of 10 micrometres on both sides, using a cell of 12 V . Calculate the energy spend by the cell in the process of deposition. If this energy is used to heat 100 g of water, calculate the rise in the temperature of the water. ECE of copper = 3 xx 10^(-7) kg C^(-1) and specific heat capacity of water = 4200 J kg^(-1) K^(-1) .

Certain amount of heat is given to 100 g of copper to increase its temperature by 21^@C . If same amount of heat is given to 50 g of water, then the rise in its temperature is (specific heat capacity of copper = 400 J kg^(-1) K^(-1) and that for water = 4200 J kg^(-1) K^(-1))

The molar heat capacity of water at constant pressure P is 75 J K ^(-1) mol ^(-1) . When 1.0 kJ of heat is supplied to 1000 g of water , which is free to expand, the increase in temperature of water is

30 gram copper is heated to increase its temperature by 20^@ C if the same quantity of heat is given to 20 gram of water the rise in its temperature. (S_(w)=4200J//kg-K & S_(cu) = 420 J//kg-K) .

RESONANCE-CALORIMETRY AND THERMAL EXPANSION-Problem
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  3. 420 J of energy supplied to 10g of water will rises its temperature by

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  15. A pendulum clock having copper rod keeos correct time at 20^0 C. It ga...

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  16. A meter scale made of steel is calibrated at 20^@C to give correct rea...

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  18. The volume occupied by a thin- wall brass vessel and the volume of a s...

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