Purchase Solution

Heat Loss and Specific Heat

Not what you're looking for?

Ask Custom Question

a) A student wishes to determine the specific heat of a metal using a coffee cup calorimeter (calorimeter constant 50.35 cal/degrees Celsius). The student heats 130.56 grams of unknown metal from 25.0 degrees Celsius to 100.4 degrees Celsius. The initial temperature of the calorimeter (coffee cup plus 30 mL water) was 23.5 degrees Celsius. The student then adds the hot metal to the coffee cup and temperature changes until equilibrium occurs. The equilibrium temperature is 30.5 degrees Celsius. Answer the following two questions:

1. How much heat is lost by the metal and what is it gained by?
2. What is the secific heat of the unknown metal?

b) A compound containg only carbon and hydrogen is 7.74% H by mass and has a molar mass of 78.1 g/mol. Determine the empirical and molecular formulae of the compound.

c) An isotope of a neutral atom has 19 electrons and 22 neutrons? Give the atomic symbol of the element.

Purchase this Solution
Solution Summary

The solution can be found by solving for the heat lost by the metal equal to the heat that the calorimeter and water absorb. In other words (deltaH of calorimeter and water) = - (deltaH of metal).

Solution Preview

1. The heat lost by the metal is gained by the water and the calorimeter.

(50.53 cal/deg C)(30.5 deg C - 23.5 deg C)
+ (30 g)(1.00 cal/g*deg C)(30.5 deg C - 23.5 deg C)
= 564 cal

2. Specific heat is how much heat something absorbs per degree Celsius per gram. In other words, J/(g*degC). Heat capacity is ...

Purchase this Solution
Free BrainMass Quizzes
Periodic table

Thermochemistry

The quiz helps in revising basic concepts about thermochemistry.

Functional groups in Organic Chemistry

You will be tested on the names of functional groups in Organic Chemistry. It is very important to know the functional groups to understand Organic reactions.

General Chemistry - Classification of Matter

This test will assess your knowledge on the classification of matter which includes elements, compounds and mixtures.

Match Elements with their Symbols

Elements are provided: choose the matching one- or two-letter symbol for each element.

View More Free Quizzes
Related BrainMass Solutions

  • Water Frozen to Ice

    This solution explains what conditions are required to freeze water in the cup to make ice using the specific heat capacity, and heat lost by the sphere, heat gained by water.

  • Change of State, calculate the heat of vaporization of water.

    Identifying given quantities by letters write: steam Ts= 100 °C and mass Ms= 27.6 g calorimeter: mass Mc = 100 g specific heat Cc= .1 cal/g °C water mass Mw = 400 g and specific heat Cw= 1 cal/ g °C water and calorimeter temp.

  • Calories and Diet

    A high specific heat is just a way of saying that the substance can absorb a lot of heat without heating up quickly. Metal conducts heat well, so it has a low specific heat as it absorbs heat and its temperature increases quickly.

  • Heat and work

    Therefore the latent heat required is 334 * m kJ The specific heat capacity of water is 4.18 kJ.kg-1.K-1.

  • If an aluminum sphere has a mass of 23.01 grams and was heat

    The final temperature of the water and the sphere was 44.0 Celsius assuming no heat was loss calculate the specific heat of the aluminum in the sphere.

  • Amount of Ice Added to Cool Water

    The general formula for heat energy gain/loss is: Where m is the mass of the substance, c is the specific heat and Tf and Ti are the final and initial temperatures respectively.

  • Heat capacity

    16303 Heat capacity A 5.00 g sample of aluminum pellets (specific heat capacity = 0.89 J/°C · g) and a 10.00 g sample of iron pellets (specific heat capacity = 0.45 J/°C · g) are heated to 100.0°C.

  • Heat Capacity and Calorimetry

    The specific heat of air at normal pressure is 1.015 j/(g.C). What is the change in air temperature, assuming the heat from the steam is all absorbed by air?

  • Heat: Specific Heat (Method of Mixtures)

    of heat is assumed to be zero, we can equate (1) and (2). 2743m = 22990 or m = 8.38 kg The specific heat for method of mixtures are examined.

  • Rate of Heat Loss through a Window

    5962 Rate of Heat Loss through a Window The two surfaces of a window are maintained at specific temperatures. The rate of heat loss through the window and the inner surface temperature are to be determined.

View More Related Solutions