150. g of hot water (c = 4.184 J/g·K) at an initial temperature of 75.00 °C is mixed with 100. g of cold water at an initial temperature of 15.00 °C. Calculate the final temperature when thermal equilibrium is reached. **Question:**150 g of hot water (c = 4.184 J/g·K) at an initial temperature of 75.00 °C is mixed with 100 g of cold water at an initial temperature of 15.00 °C. Calculate the final temperature when thermal equilibrium is reached.**Options:**- 51.0 °C- 25.5 °C- 45.0 °C- 39.0 °C**Explanation:**This question involves calculating the equilibrium temperature when two masses of water at different temperatures are mixed. The specific heat capacity of water is given (4.184 J/g·K).To solve, apply the principle of conservation of energy, where the heat lost by the hot water will equal the heat gained by the cold water. The formula used is:\[ m_1 \times c \times (T_i - T_f) = m_2 \times c \times (T_f - T_i) \]Where:- \( m_1 \) and \( m_2 \) are the masses of the hot and cold water, respectively.- \( T_i \) and \( T_f \) are the initial and final temperatures, respectively.- \( c \) is the specific heat capacity.Substitute the given values to find the final temperature \( T_f \).

Given that: Mass of hot water = 150.0 g Initial temperature of hot water = 75°C Mass of cold water =…

150. g of hot water (c = 4.184 J/g-K) at an initial temperature of 75.00 °C is mixed with 100. g of cold water at an initial temperature of 15.00 °C. Calculate the final temperature when thermal equilibrium is reached. O 51.0 °C 25.5 °C O 45.0 °C O 39.0 °C

150. g of hot water (c = 4.184 J/g-K) at an initial temperature of 75.00 °C is mixed with 100. g of cold water at an initial temperature of 15.00 °C. Calculate the final temperature when thermal equilibrium is reached. O 51.0 °C 25.5 °C O 45.0 °C O 39.0 °C

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

Define specific heat capacity. the quantity of heat required to raise the temperature of 1 mole of a substance by 1°C the quantity of heat required to change a system's temperature by 1°C the quantity of heat required to raise the temperature of 1 gram of a substance by 1°C the quantity of heat required to raise the temperature of 1 gram of a substance by 1°F the quantity of heat required to raise the temperature of 1 liter of a substance by 1°C
A 5.71 g sample of an unknown salt (MM = 116.82 g/mol) is dissolved in 150.00 g water in a coffee cup calorimeter. Before placing the sample in the water, the temperature of the salt and water is 23.72 °C. After the salt has completely dissolved, the temperature of the solution is 28.54 °C. What is the enthalpy change (in kJ/mol of salt) for the dissolution reaction?
Could you please assist me with some answers to my questions?
In a coffee cup calorimeter, 50.0 mL of 1.00 M NaOH and 50.00 mL of 1.00 M HCl are mixed. Both solutions were originally at 24.6 C. After the reaction, the final temperature is 31.3 C. Given that the density of NaCl solution is 1.038 g/mL and the specific heat of NaCl solution is 3.87, calculate the change in enthalpy of neutralization per mole for the reaction of HCl with NaOH. Assume that no heat is lost to the surroundings.
22. A 44.97−g sample of water at 72.2°C is added to a sample of water at 25.7°C in a constant-pressure calorimeter. If the final temperature of the combined water is 38.4°C and the heat capacity of the calorimeter is 26.3 J/°C, calculate the mass of the water originally in the calorimeter.
A 3.000 g sample of food was then burned in a calorimeter to determine its Calories per gram content. 50mL of water were used to determine ΔT of the reaction. Initial temperature of the water at the start of the combustion reaction was 20oC, final temperature was 99oC. Calorimeter constant for this calorimeter is 10.0J/oC. How much heat has been absorbed by the water sample?
100.0 mL of 1.00 M HBr at 22.5 °C is combined with 100.0 mL of 1.00 M LiOH at 22.5 °C. The mixture reaches a temperature of 29.2 °C. Assume the density of the solutions is 1.00 g/mL and their specific heat capacity is the same as pure water. What is the molar heat of reaction?
3.76 g of MgSO4 is placed into 100.0 mL of water. The water's temperature increases by 6.70 °C. Calculate AH, in kJ/mol, for the dissolution of MgSO4. (The specific heat of water is 4.184 J/g °C and the density of the water is 1.00 g/ mL). You can assume that the specific heat of the solution is the same as that of water.
The molar heat of solution of a substance is found to be +21.38 kJ/mol. The addition of 0.100 mol of this substance to 1.000L of water initially at 40.0 degrees celsius results in a temperature decrease. Assume the specific heat of the resulting solution to be equal to that of pure water. Find the final temperature of the solution (Also assume that the heat capacity of the calorimeter is negligible).
A student determines the heat of dissolution of solid potassium bromide using a coffee-cup calorimeter of negligible heat capacity.When 4.93 g of KBr(s) is dissolved in 108.00 g of water, the temperature of the solution drops from 25.00 to 23.11 °C. Based on the student's observation, calculate the enthalpy of dissolution of KBr(s) in kJ/mol. Assume the specific heat of the solution is 4.184 J/g°C.ΔHdissolution = kJ/mol
When a 6.00 g sample of KSCN is dissolved in water in a calorimeter that has a total heat capacity of 3.74 kJ · K¯¹, the temperature decreases by 0.400 K. Calculate the molar heat of solution of KSCN. ΔΗsoln = kJ/mol