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Chapter 2: Problem 156

When \(10 \mathrm{~kJ}\) of heat energy is added to a beaker containing \(250 \mathrm{~g}\) of water initially at \(23.0{ }^{\circ} \mathrm{C}\), what is the final temperature of the water?

Short Answer

Expert verified
The final temperature of the water when 10 kJ of heat energy is added to the beaker is approximately \(32.57^{\circ}\)C.

Step by step solution

01

Identify the given information

We are given the following information: - Heat energy added (Q) = 10 kJ (convert to J by multiplying by 1000, so Q = 10000 J) - Mass of water (m) = 250 g (convert to kg by dividing by 1000, so m = 0.25 kg) - Initial temperature (T1) = 23.0 °C - Specific heat capacity of water (c) = 4.18 kJ/kg·°C = 4180 J/kg·°C We need to find the final temperature (T2).
02

Use the heat transfer formula

The formula for calculating the heat transfer in a substance is: Q = mcΔT where ΔT = T2 - T1
03

Rearrange the formula to solve for the final temperature T2

To find the final temperature T2, we need to rearrange the formula as follows: ΔT = Q / (mc) T2 = T1 + Q / (mc)
04

Plug in the given values and solve for T2

Now plug in the given information into the formula and calculate the final temperature: T2 = 23.0 + (10000 J) / (0.25 kg × 4180 J/kg·°C) T2 = 23.0 + (10000) / (1045) T2 = 23.0 + 9.57 ≈ 32.57 °C The final temperature of the water when 10 kJ of heat energy is added to the beaker is approximately 32.57 °C.

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