Answer:
There is 98.76 kJ energy transfered to the water as heat.
Explanation:
<u>Step 1:</u> Data given
Mass of copper bowl = 110 grams
Mass of water = 240 grams
Temperature of water and copper = 21.0 °C
Mass of the hot copper cylinder = 410 grams
8.6 grams being converted to steam
Final temperature = 100 °C
<u>Step 2:</u> Calculate the energy gained by the water:
Q = m(water)*C(water)*ΔT + m(vapor)*Lw
⇒with mass of water = 0.240 kg
⇒ with C(water) = the heat capacity of water = 4184 J/kg°C
⇒ with ΔT = the change in temperature = T2 - T1 = 100 °C - 21.0 = 79°C
⇒ with mass of vapor = 8.60 grams = 0.0086 kg
⇒ with Lw = The latent heat of vaporization (water to steam) = 22.6 *10^5 J/kg
Q = 0.24kg * 4184 J/kg°C *79°C + 0.0086 kg*22.6*10^5 J/kg
Q = 79328.64 + 19436 = 98764.64 J = 98.76 kJ
There is 98.76 kJ energy transfered to the water as heat.
