Calculate the energy required to heat 1.10 kg of water from 49.1 °C to 62.6 °C. Assume the specific heat capacity of water under these conditions is 4.18 J-g ¹.K-¹ . Round your answer to 3 significant digits. 1 H 0.0 X Olo $ A
Thermochemistry
Thermochemistry can be considered as a branch of thermodynamics that deals with the connections between warmth, work, and various types of energy, formed because of different synthetic and actual cycles. Thermochemistry describes the energy changes that occur as a result of reactions or chemical changes in a substance.
Exergonic Reaction
The term exergonic is derived from the Greek word in which ‘ergon’ means work and exergonic means ‘work outside’. Exergonic reactions releases work energy. Exergonic reactions are different from exothermic reactions, the one that releases only heat energy during the course of the reaction. So, exothermic reaction is one type of exergonic reaction. Exergonic reaction releases work energy in different forms like heat, light or sound. For example, a glow stick releases light making that an exergonic reaction and not an exothermic reaction since no heat is released. Even endothermic reactions at very high temperature are exergonic.
![**Problem Statement:**
Calculate the energy required to heat 1.10 kg of water from 49.1 °C to 62.6 °C. Assume the specific heat capacity of water under these conditions is \(4.18 \ \text{J} \cdot \text{g}^{-1} \cdot \text{K}^{-1}\). Round your answer to 3 significant digits.
**Input Fields:**
1. **Blank Input Field:** This signifies an empty block for the user to enter their response.
2. **Option Buttons:**
- \( \Box \)
- \( \Box \right\Box \right\Box x10 \)
- \( \Box \)
- \( \Box x \)
- \( \Box \mu \)
These option buttons likely represent different units or multipliers that the user can select from to enter their final answer in the correct format.
**Solution Method:**
To find the energy (\(Q\)) required to heat the water, use the formula:
\[ Q = m \cdot c \cdot \Delta T \]
where:
- \(m\) is the mass of the water in grams
- \(c\) is the specific heat capacity (\(4.18 \ \text{J} \cdot \text{g}^{-1} \cdot \text{K}^{-1}\))
- \(\Delta T\) is the change in temperature in °C (or K, since the difference in temperature is the same in both units)
Given:
- Mass \(m\) = 1.10 kg = 1100 g (conversion from kg to g)
- Initial temperature \(T_i\) = 49.1 °C
- Final temperature \(T_f\) = 62.6 °C
- Change in temperature \(\Delta T\) = \(T_f - T_i = 62.6 °C - 49.1 °C = 13.5 \ °C\)
Substitute the values into the formula:
\[ Q = 1100 \ g \times 4.18 \ \frac{\text{J}}{\text{g} \cdot \text{K}} \times 13.5 \ \text{K} \]
\[ Q = 1100 \times 4.18 \times 13.5 \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2869f911-fc69-4f3e-bbf4-69a25f43f41f%2Ff87ce1bd-ad08-41a8-a37f-f25507a8f3ee%2Fb4g7cax_processed.jpeg&w=3840&q=75)
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