Question

If \(125.6 \mathrm{kJ}\) of heat are supplied to \(5.00 \times 10^{2} \mathrm{g}\) of water at \(22^{\circ} \mathrm{C},\) what is the final temperature of the water?

Short answer

Answer: The final temperature of the water is approximately 81.88°C.

Step-by-step solution

Convert the heat energy from kJ to J

Since 1 kJ = 1000 J, convert 125.6 kJ to J: 125.6 kJ × 1000 = 125600 J

Calculate the heat absorbed by the water

To calculate the heat absorbed by the water, we use the formula: Q = mcΔT Where Q is the heat absorbed, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature. We are given the mass of water and its initial temperature. We also know that the specific heat capacity of water is about 4.18 J/g°C. We will solve for ΔT to find the final temperature: ΔT = Q / (mc)

Substitute the given values into the formula

Substitute the values into the formula with m = 5.00 x 10^2 g, c = 4.18 J/g°C, and Q = 125600 J: ΔT = 125600 J / ( (5.00 x 10^2 g) x 4.18 J/g°C)

Calculate the change in temperature (ΔT)

Perform the calculations: ΔT ≈ 59.88°C

Determine the final temperature of the water

To find the final temperature, add the change in temperature (ΔT) to the initial temperature (22°C): Final temperature = Initial temperature + ΔT Final temperature = 22°C + 59.88°C Final temperature ≈ 81.88°C The final temperature of the water is approximately 81.88°C.