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Working with specific heat capacity

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Question: A piece of glass has a temperature of 85.4C. Liquid that has a temperature of 43.0C is poured over the glass, completely covering it, and the temperature at equilibrium is 51.4C. The mass of the glass and the liquid is the same. Ignoring the container that holds the glass and liquid and assuming that the heat lost to or gained from the surroundings is negligible, determine the specific heat capacity of the liquid.

I know that the equation needed is:
Q=cm(delta T)
where Q=the heat that must be supplied or removed to change the temperature of a substance
and c=specific heat capacity
and m=mass of substance
and delta T= the change in temperature

I also know that in order to find the mass of the glass and liquid that i must plug in the necessary information into the equation with the specific heat capacity of glass and solve for m. Then I must plug the mass into the equation for the liquid and solve for c.

The problem that I am experiencing is how to determine Q.

This is what I have tried and is NOT correct:
(Glass)
34=(840)m(42.4)
m=.001 kg
(Liquid)
8.1=c(.001)(42.4)
c= 191.038 J/kgC

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Solution Preview

Quantity of Heat lost = mass * specific heat * dT

If glass loses a particular quantity of heat then the liquid gains that much amout of heat enery.

Thus we will write, Mg * Cg * dT_forglass = Ml * Cl * dT_for liquid

...

Solution Summary

Solution gives all mathematical steps.

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A crate of fruit with a mass of 35.0 kg and a specific heat capacity of 3800 J/kg.K slides 8.30m down a ramp inclined at an angle of 40.0 degrees below the horizontal (acceleration due to gravity is 9.81m/s^2).

If the crate was at rest at the top of the incline and has a speed of 2.40m/s at the bottom, the work (W_f) done on the crate, by friction is -1730J (The frictional force opposes the motion of the crate, so the work done on the crate by friction must be a negative quantity).

If an amount of heat equal to the magnitude of the work done by friction is absorbed by the crate of fruit and the fruit reaches a uniform final temperature, what is its temperature change ("deltaT") ?

"deltaT" = ___________________________________________ oC

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