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°
Problems 1. To what temperature will 4500 J of work raise 1.0 kg of water initially...
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