1 Cal = 4180 J.
Q = ΔU1 + ΔU2 =
=c1•m1 •ΔT + c2•m2 •ΔT =
= ΔT •(c1•m1 + c2•m2 ) =
= 95• (4180•0.115•100 + 4180•200)= =8.4•10^7 J
A 100 kg steam boiler is made of steel and contains 200 kg of water at 5.00°C. How much heat is required to raise the temperature of both the boiler and water to 100°C?
The specific heat of Steel is 0.115kcal/kg°C. The specific heat of Water is 1.00kcal/kg°C
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