Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Problem 33

What mass of water at \(25.0^{\circ} \mathrm{C}\) added to a Styrofoam cup containing two 50.0 -g ice cubes from a freezer at \(-15.0^{\circ} \mathrm{C}\) will result in a final temperature of \(5.0^{\circ} \mathrm{C}\) for the drink?

Short Answer

Expert verified
Answer: 1648 g

Step by step solution

01

Calculate the heat required to raise the temperature of ice cubes to 0°C

We first need to calculate the heat required to raise the temperature of both ice cubes (-15.0°C) to 0°C. We will use the equation \(Q = mcΔT\): \(Q = (m_{ice})(c_{ice})(ΔT_{ice})\) Here, \(m_{ice} = 2 \times 50.0\,\text{g} = 100.0\,\text{g}\) (mass of both ice cubes), \(c_{ice} = 2.093\,\frac{J}{g°C}\) (specific heat capacity of ice), and \(ΔT_{ice} = 15.0\,°\text{C}\) (change in temperature from -15.0°C to 0°C). \(Q_{ice} = (100.0\,\text{g})(2.093\,\frac{J}{g°C})(15.0\,°\text{C}) = 3140\,\text{J}\) (rounded to 3 significant figures)
02

Calculate the heat required to change ice from solid to liquid

Next, we need to calculate the heat required to change the solid ice cubes into liquid water at 0°C. We will use the equation \(Q = mL\): \(Q = (m_{ice})(L_{fusion})\) Here, \(m_{ice} = 100.0\,\text{g}\) (mass of both ice cubes) and \(L_{fusion} = 334\,\frac{J}{g}\) (latent heat of fusion for water). \(Q_{fusion} = (100.0\,\text{g})(334\,\frac{J}{g}) = 33400\,\text{J}\)
03

Calculate the heat required to raise the temperature of liquid water to the final temperature

Now, we need to calculate the heat required to raise the temperature of the entire mixture (100 g of liquid water and the mass of water) to the final temperature of 5.0°C. We will use the equation \(Q = mcΔT\): \(Q = (m_{water}+m_{ice})(c_{water})(ΔT_{water})\) Here, \(m_{water} + m_{ice} = 100.0\,\text{g} + m_{water}\) (mass of water added plus the mass of ice cubes), \(c_{water} = 4.18\,\frac{J}{g°C}\) (specific heat capacity of water), and \(ΔT_{water} = 5.0\,°\text{C}\) (change in temperature from 0°C to 5.0°C). \(Q_{water} = (100.0\,\text{g} + m_{water})(4.18\,\frac{J}{g°C})(5.0\,°\text{C}) = (100.0\,\text{g} + m_{water})\times 20.9\,\text{J} \)
04

Equate the heat lost by the water to the heat gained by the ice

The heat lost by the water (\(Q_{water}\)) will be equal to the heat gained by the ice (\(Q_{total}\)), where \(Q_{total} = Q_{ice} + Q_{fusion}\): \(Q_{water} = Q_{total}\) \((100.0\,\text{g} + m_{water})\times 20.9\,\text{J} = 3140\,\text{J} + 33400\,\text{J}\)
05

Solve for the mass of water

Now, we can solve for the mass of water (\(m_{water}\)) by simplifying the equation from Step 4: \((100.0\,\text{g} + m_{water})\times 20.9\,\text{J} = 36540\,\text{J}\) Divide both sides by 20.9 J: \(100.0\,\text{g} + m_{water} = 1748\,\text{g}\) (rounded to 3 significant figures) Subtract 100.0 g from both sides: \(m_{water} = 1648\,\text{g}\) Therefore, 1648 g of water at 25.0°C needs to be added to the two 50.0-g ice cubes to achieve a final temperature of 5.0°C for the drink.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A heating coil inside an electric kettle delivers \(2.1 \mathrm{kW}\) of electric power to the water in the kettle. How long will it take to raise the temperature of \(0.50 \mathrm{kg}\) of water from \(20.0^{\circ} \mathrm{C}\) to \(100.0^{\circ} \mathrm{C} ?\) (tutorial: heating)
If a leaf is to maintain a temperature of \(40^{\circ} \mathrm{C}\) (reasonable for a leaf), it must lose \(250 \mathrm{W} / \mathrm{m}^{2}\) by transpiration (evaporative heat loss). Note that the leaf also loses heat by radiation, but we will neglect this. How much water is lost after 1 h through transpiration only? The area of the leaf is \(0.005 \mathrm{m}^{2}\).
You are given \(250 \mathrm{g}\) of coffee (same specific heat as water) at \(80.0^{\circ} \mathrm{C}\) (too hot to drink). In order to cool this to $\left.60.0^{\circ} \mathrm{C}, \text { how much ice (at } 0.0^{\circ} \mathrm{C}\right)$ must be added? Ignore heat content of the cup and heat exchanges with the surroundings.
For a cheetah, \(70.0 \%\) of the energy expended during exertion is internal work done on the cheetah's system and is dissipated within his body; for a dog only \(5.00 \%\) of the energy expended is dissipated within the dog's body. Assume that both animals expend the same total amount of energy during exertion, both have the same heat capacity, and the cheetah is 2.00 times as heavy as the dog. (a) How much higher is the temperature change of the cheetah compared to the temperature change of the dog? (b) If they both start out at an initial temperature of \(35.0^{\circ} \mathrm{C},\) and the cheetah has a temperature of \(40.0^{\circ} \mathrm{C}\) after the exertion, what is the final temperature of the dog? Which animal probably has more endurance? Explain.
A birch tree loses \(618 \mathrm{mg}\) of water per minute through transpiration (evaporation of water through stomatal pores). What is the rate of heat lost through transpiration?
See all solutions

Recommended explanations on Physics Textbooks

Circular Motion and Gravitation

Read Explanation

Radiation

Read Explanation

Engineering Physics

Read Explanation

Fields in Physics

Read Explanation

Oscillations

Read Explanation

Translational Dynamics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free