/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 43 You are given \(250 \mathrm{g}\)... [FREE SOLUTION] | 91影视

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Chapter 14: Problem 43

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.

Short Answer

Expert verified
Answer: Approximately \(35.7 \mathrm{g}\) of ice is needed to cool the coffee to \(60.0^{\circ} \mathrm{C}\).

Step by step solution

01

Write the energy conservation equation

The energy conservation equation can be written as: heat gained by the ice + heat gained by the melted ice = heat lost by the coffee
02

Write the formula for heat gained/lost

Heat gained/lost can be written as \(Q = mc螖T\) for a given substance, where \(Q\) is the heat, \(m\) is the mass, \(c\) is the specific heat capacity, and \(螖T\) is the change in temperature. In this case, we have three substances: ice, melted ice (water), and coffee.
03

Introduce the latent heat of fusion for ice

During ice melting, we must account for the amount of heat absorbed by ice to turn to water. The heat obtained from the coffee will first be used to melt the ice; the melted ice will then gain heat from the remaining heat available. The latent heat of fusion can be written as \(Q_{fusion} = mL\), where \(Q_{fusion}\) is the heat absorbed during the phase change, \(m\) is the mass of ice, and \(L\) is the latent heat of fusion.
04

Write the energy conservation equation with known values and variables

Considering all the steps above, we can write the equation for this problem, taking \(m_{ice}\) as the mass of ice added: \((m_{ice}L) + (m_{ice}c_w螖T_{melted\_ice}) = (m_{coffee}c_w螖T_{coffee})\) where \(c_w = 4.18 \mathrm{J/g\:C}\) is the specific heat capacity of water, \(L = 334 \mathrm{J/g}\) is the latent heat of fusion for water, \(m_{coffee} = 250 \mathrm{g}\), \(螖T_{melted\_ice} = (60 - 0)\), and \(螖T_{coffee} = (80 - 60)\).
05

Solve the equation for mass of ice

Simplify and plug in known values: \(m_{ice}(334) + m_{ice}(4.18)(60) = (250)(4.18)(20)\) \(m_{ice}(334 + 250.8) = (250)(83.6)\) \(m_{ice} = \frac{(250)(83.6)}{584.8}\) \(m_{ice} \approx 35.7 \mathrm{g}\) So, approximately \(35.7 \mathrm{g}\) of ice is needed to cool the coffee to \(60.0^{\circ} \mathrm{C}\).

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Most popular questions from this chapter

A thermometer containing \(0.10 \mathrm{g}\) of mercury is cooled from \(15.0^{\circ} \mathrm{C}\) to \(8.5^{\circ} \mathrm{C} .\) How much energy left the mercury in this process?

Imagine a person standing naked in a room at \(23.0^{\circ} \mathrm{C}\) The walls are well insulated, so they also are at \(23.0^{\circ} \mathrm{C}\) The person's surface area is \(2.20 \mathrm{m}^{2}\) and his basal metabolic rate is 2167 kcal/day. His emissivity is \(0.97 .\) (a) If the person's skin temperature were \(37.0^{\circ} \mathrm{C}\) (the same as the internal body temperature), at what net rate would heat be lost through radiation? (Ignore losses by conduction and convection.) (b) Clearly the heat loss in (a) is not sustainable-but skin temperature is less than internal body temperature. Calculate the skin temperature such that the net heat loss due to radiation is equal to the basal metabolic rate. (c) Does wearing clothing slow the loss of heat by radiation, or does it only decrease losses by conduction and convection? Explain.
If \(4.0 \mathrm{g}\) of steam at \(100.0^{\circ} \mathrm{C}\) condenses to water on a burn victim's skin and cools to \(45.0^{\circ} \mathrm{C},\) (a) how much heat is given up by the steam? (b) If the skin was originally at $37.0^{\circ} \mathrm{C},$ how much tissue mass was involved in cooling the steam to water? See Table 14.1 for the specific heat of human tissue.
Bare, dark-colored basalt has a thermal conductivity of 3.1 $\mathrm{W} /(\mathrm{m} \cdot \mathrm{K}),$ whereas light-colored sandstone's thermal conductivity is only \(2.4 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K}) .\) Even though the same amount of radiation is incident on both and their surface temperatures are the same, the temperature gradient within the two materials will differ. For the same patch of area, what is the ratio of the depth in basalt as compared with the depth in sandstone that gives the same temperature difference?
How much heat is required to raise the body temperature of a \(50.0-\mathrm{kg}\) woman from \(37.0^{\circ} \mathrm{C}\) to $38.4^{\circ} \mathrm{C} ?$
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