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334.8 Joules
This heat is 32,48 joules.
specific heat(; your welcome!
The heat required will be equal to 87.8 kJ.Energy = Specific heat capacity x mass x change in TemperatureQ = Cg x m x change T= 4.18 x 300 x 70= 87780 J= 87.8 kJDensity of water is 1.00 g / cm3Heat capacity of water = 4.18 J oC-1g-1
q = (20.5 g)(0.21 J/gC)(230o C- 30o C) = 861 Joules =========
334.8 Joules
Specific heat of aluminum is 0.902 J/gC Use this formula. q(Joules) = mass * specific heat * change in temperature q = 106 grams Al * 0.902 J/gC *(121 C = 96 C) = 2390 Joules of heat
539 joules, approx.
The needed heat is 47,65 Joules.
This heat is 32,48 joules.
Heat energy Q = mass x specific heat capacity x temperature change. Q = m*c*delta T Q = Joules m = kg c (aluminum) = 895.8 J/kg delta T = degr.C temp. change Answer: Q = (20/1000) x 895.8 x 5 = 89.58 Joules (Specific heat capacity of aluminum is obtained by multiplying its specific heat of 0.214 with c of water which is 4186 J/kg = 0.214 x 4186 = 895.8 J/kg).
0.902 J/gC is the specific heat of aluminum. Use (q)heat(Joules) = mass * specific heat * change in temperature q = (480 g)(0.902 J/gC)(243 C - 23 C) = 95261.2 Joules ------------------------------( call it 9.5 X 104 Joules )
How much heat (in calories) is required to heat a 43 g sample of aluminum from 72 F to 145F
specific heat(; your welcome!
specific heat(; your welcome!
The heat required will be equal to 87.8 kJ.Energy = Specific heat capacity x mass x change in TemperatureQ = Cg x m x change T= 4.18 x 300 x 70= 87780 J= 87.8 kJDensity of water is 1.00 g / cm3Heat capacity of water = 4.18 J oC-1g-1
The energy in Joules required to cause the 200 K increase in the temperature of 3.5 kg of Aluminium (with a specific heat capacity or SHC of 897 J kg K) equals:Energy (J) = Change in Temp (K) x mass (kg) x SHC (J kg K)Energy (J) = 200 x 3.5 x 897= 627,900 Joules.