A 110 g copper bowl contains 240 g of water, both at 21.0°C. A very hot 410 g copper cylinder is dropped into the water, causing the water to boil, with 8.60 g being converted to steam. The final temperature of the system is 100°C. Neglect energy transfers with the environment. (a) How much energy is transferred to the water as heat

There is 98.76 kJ energy transfered to the water as heat.

Explanation:

Step 1: 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

Step 2: 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.