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Problems 1. To what temperature will 4500 J of work raise 1.0 kg of water initially at 10 degrees Celsius? 2. How much work (

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Answer #1

1)Solution-

The specific heat of water = 4186 J/kg*C°
Now 4186 Joules of heat energy will increase the temperature of 1 kg of water 1°C

Now the heat energy to increase temperature = mass * Sp.Ht * ∆T
4500 = 1.0 * 4186 * ∆T
=>∆T = 4500 / (1.0 * 4186) = 1.075°

Final temperature = 10.0 + 1.075

= 11.075°

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