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Chapter 6: Problem 60

A biology experiment requires the preparation of a water bath at $37.0^{\circ} \mathrm{C}$ (body temperature). The temperature of the cold tap water is \(22.0^{\circ} \mathrm{C},\) and the temperature of the hot tap water is \(55.0^{\circ} \mathrm{C} .\) If a student starts with 90.0 \(\mathrm{g}\) cold water, what mass of hot water must be added to reach $37.0^{\circ} \mathrm{C} ?$

Short Answer

Expert verified
To reach a water bath temperature of \(37.0^{\circ}\mathrm{C}\), a student must add 75.0 g of hot water to the initial 90.0 g of cold water.

Step by step solution

01

Understand the Heat Exchange Principle

To reach a common temperature, the heat gained by the cold water must equal the heat lost by the hot water. We can use the specific heat formula to express this heat exchange: Heat gained by cold water = Heat lost by hot water
02

Write the Formula for the Specific Heat of Water

The specific heat formula is given as: Q = mcΔT Where: - Q is the heat absorbed or released; - m is the mass of the substance; - c is the specific heat capacity, which for the water is 4.18 J/g°C; and - ΔT is the change in temperature.
03

Write the Equations for Cold and Hot Water

For cold water, the mass is given as 90.0 g. Let's denote the mass of hot water added as m. Then, the equations for heat gained by cold water (Q_c) and heat lost by hot water (Q_h) can be written as: Q_c = (90.0 g)(4.18 J/g°C)(37.0°C - 22.0°C) Q_h = (m)(4.18 J/g°C)(55.0°C - 37.0°C) As stated in Step 1, Q_c = Q_h.
04

Solve for the Mass of Hot Water

Equate the two equations and solve for m: (90.0 g)(4.18 J/g°C)(37.0°C - 22.0°C) = (m)(4.18 J/g°C)(55.0°C - 37.0°C) Simplify and cancel like terms: (90.0 g)(15.0°C) = (m)(18.0°C) Divide both sides by 18.0°C: m = \(\frac{(90.0 g)(15.0°C)}{(18.0°C)}\) m = 75.0 g
05

Final Answer

Therefore, to reach a water bath temperature of \(37.0^{\circ}\mathrm{C}\), a student must add 75.0 g of hot water to the initial 90.0 g of cold water.

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Most popular questions from this chapter

Calculate \(q, w, \Delta E,\) and \(\Delta H\) for the process in which 88.0 g of nitrous oxide (laughing gas, \(\mathrm{N}_{2} \mathrm{O} )\) is cooled from \(165^{\circ} \mathrm{C}\) to \(55^{\circ} \mathrm{C}\) at a constant pressure of 5.00 \(\mathrm{atm} .\) The molar heat capacity for $\mathrm{N}_{2} \mathrm{O}(g)\( is 38.7 \)\mathrm{J} /^{\prime} \mathrm{C} \cdot$ mol.

Explain why \(\Delta H\) is obtained directly from coffee-cup calorimeters, whereas \(\Delta E\) is obtained directly from bomb calorimeters.

The enthalpy change for the reaction $$ \mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ is \(-891 \mathrm{kJ}\) for the reaction as written. a. What quantity of heat is released for each mole of water formed? b. What quantity of heat is released for each mole of oxygen reacted?

In a coffee-cup calorimeter, 100.0 \(\mathrm{mL}\) of 1.0 \(\mathrm{M}\) NaOH and 100.0 \(\mathrm{mL}\) of 1.0 \(\mathrm{M} \mathrm{HCl}\) are mixed. Both solutions were originally at \(24.6^{\circ} \mathrm{C}\) . After the reaction, the final temperature is \(31.3^{\circ} \mathrm{C}\) . Assuming that all the solutions have a density of 1.0 \(\mathrm{g} / \mathrm{cm}^{3}\) and a specific heat capacity of \(4.18 \mathrm{J} / \mathrm{C} \cdot \mathrm{g},\) calculate the enthalpy change for the neutralization of \(\mathrm{HCl}\) by NaOH. Assume that no heat is lost to the surroundings or to the calorimeter.

A serving size of six cookies contains 4 g of fat, 20 of carbohydrates, and 2 g of protein. If walking 1.0 mile consumes 170 kJ of energy, how many miles must you walk to burn off enough calories to eat six cookies? Assume the energy content of fats, carbohydrates, and proteins are 8 kcallg, 4 kcallg, and 4 kcallg, respectively.
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