# Heat Transfer Problem

## Homework Statement

What mass of water at 25dC (d = degrees) must be allowed to come to thermal equalibrium with a 1.85kg cube of aluminum initially at 1.5*10*2dC to lower the temperature of the aluminum to 65dC? Assume the water doesn't turn to steam.

## Homework Equations

Q = Del(T)m(C), DelT= change in temperature, m = mass, C = specific heat. It doesn't actually give any equation I just assume that it deals with that one.

## The Attempt at a Solution

(after conversion to Kelvin)

85K * (1.85kg) * (.91kJ/(kgK)aluminum = 143.0975kJ

143.0975kJ = 125K * m * (4.18kJ/kgK)water

I assumed this was the amount energy involved in the temperature change from aluminum at 423K to 338K, and assumed the energy is conserved since conservation of energy was a topic recently and so assumed the amount of energy that the aluminum cooled by the was amount of energy the water gained. It makes sense at first me at first except I don't understand exactly how to consider the "equilibrium" part of it, I know from calculus there's Newton's law of Cooling and so I don't know why/how that's used with varying temperatures and masses to cool something to an equilibrium instead, and neither do I get exactly how an equilibrium is being considered in this situation. Anyway, it said the answer was .845kg and I got .273kg.

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Why does everyone get their question answered but me :(

The final temperature for the water will equal the final temperature for the aluminum, they're in thermal equilibrium.

You listed change in temperature for water as going from initial to aluminums initial, 150. Sub in delta T to be 40 instead of 125.

SteamKing
Staff Emeritus
Homework Helper
'Thermal equilibrium' means the water and the aluminum have reached the same temperature. As heat is transferred from the aluminum cube to the water, the metal cools while the water heats up.

You aren't being asked how long this process takes, so Newton's Law of Cooling isn't useful for this problem.

Although specific heats are given in units of kJ/K-kg and temperatures are given in degrees C, you don't need to convert temperature from degrees C to degrees K because you are interested in the amount of heat it takes to change the temperature of a substance, and a temperature difference of 1 degree C is the same temperature difference of 1 degree K.

It's not clear why you said the temperature of the water would increase by 125 degrees K. After all, since the water had an initial temperature of 25 C, an increase of 75 C would bring the water to its boiling point of 100 C. That's why your mass calculation failed.

I didn't get the exact number but I got pretty close, thanks.