Given,
amount of heat Q = 3.6 * 10 6J
mass of water m = 50 kg
specific heat capacity s = 4200 J/ kg°C
difference in temp dt = ??
We have ,
\(DT\;=\;\frac Q{M\;\ast\;S\;}\)
dt = 17.14 °C
Final temperature is 17.14 + 30 = 47.14°C
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What will be the temp raised for 50 kg water with 30°C , power 1000 W giving 3.6 * 10 ^ 6 J.
Given,
amount of heat Q = 3.6 * 10 6J
mass of water m = 50 kg
specific heat capacity s = 4200 J/ kg°C
difference in temp dt = ??
We have ,
\(DT\;=\;\frac Q{M\;\ast\;S\;}\)
dt = 17.14 °C
Final temperature is 17.14 + 30 = 47.14°C
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