/*! 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} Q40P Calculate the specific heat of a... [FREE SOLUTION] | 91影视

91影视

Calculate the specific heat of a metal from the following data. A container made of the metal has a mass of 3.6kgand contains14kgof water. A1.8kgpiece of the metal initially at a temperature of180掳颁is dropped into the water. The container and water initially have a temperature of16.0掳颁, and the final temperature of the entire (insulated) system is18.0掳颁.

Short Answer

Expert verified

The specific heat of the metal is0.411kJ/(kg.K)

Step by step solution

01

The given data

i) Mass of water(Mw)=14鈥塳驳

ii) Mass of metal(Mm)=1.8鈥塳驳

iii) Another mass(Mc)ofmetal=3.6鈥塳驳

iv) Initial temperature(Ti1)=1800C

v) Initial temperature(Ti2)=160C

vi) Final temperature of the entire system(Tf)=180C

02

Understanding the concept of calorimetry

Using the concept of calorimetry, i.e., energy lost by a hot object is equal to the energy gained by the cold object when they reach equilibrium, we can write the equation for energy lost or gained by hot and cold objects respectively. We can use the formula for specific heat to write the equation for heat lost or heat gained in terms of mass, specific heat, and difference in the temperature. This equation can be solved to find the specific heat of the given material.

Formula:

The heat energy required by a body, 鈥(i)

Where,m = mass

c= specific heat capacity

T= change in temperature

Q= required heat energy

03

Calculation of the specific heat of the metal

We can write the above formula as simplified for equilibrium using equation (i) as given:

(Mwcw+Mccm)(TfTi2)+Mmcm(TfTi1)=0

By solving forwe get the above formula as:

cm=Mwcw(Ti2Tf)Mc(TfTi2)+Mm(TfTi1)=(14鈥塳驳)(4.18鈥塳闯kg.K)(160C180C)3.6鈥塳驳(180C160C)+1.8kg(180C1800C)=117.04284.4kJkg.K=0.411kJkg.K

Hence, the value of the specific heat is0.411kJkg.K

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

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

The average rate at which energy is conducted outward through the ground surface in North America is54.0mW/m2, and the average thermal conductivity of the near-surface rocks is2.50W/mK. Assuming a surface temperature of10.0C, find the temperature at a depth of35.0km(near the base of the crust). Ignore the heat generated by the presence of radioactive elements.

A small electric immersion heater is used to heat 100gof water for a cup of instant coffee. The heater is labeled 鈥200watts鈥 (it converts electrical energy to thermal energy at this rate). Calculate the time required to bring all this water from23.0Cto100C, ignoring any heat losses.

A cylindrical copper rod of length1.2mand cross sectional area4.8鈥塩尘2is insulated to prevent heat loss through its surface. The ends are maintained at a temperature difference of1000Cby having one end in a water 鈥 ice mixture and the other in a mixture of boiling water and steam.(a) At what rate is energy conducted along the rod? (b) At what rate does ice melt at the cold end?

A sample of gas expands fromV1=1.0m3 andp1=40Pa toV2=4.0m3 andp2=10Paalong path Bin the p-Vdiagram in Fig. 18-58. It is then compressed back to V1 along either path Aor path C. Compute the net work done by the gas for the complete cycle along (a) path BAand (b) path BC.

Figure 18-26 shows three different arrangements of materials 1, 2, and 3 to form a wall. The thermal conductivities are k1>k2>k3. The left side of the wall is20 higher than the right side. Rank the arrangements according to (a) the (steady state) rate of energy conduction through the wall and (b) the temperature difference across material 1, greatest first.

See all solutions

Recommended explanations on Physics Textbooks

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.