In grinding a steel knife blade (specific heat = 0.11 cal/g-c),the metal can get as hot as 400c. if the blade has a mass of 80g, what is the minimum amount of water needed at 20c if the water isnot to rise above the boiling point when the hot blade is quenchedin it?

33 g.

Explanation:

Assuming no heat transfer can be possible except for heat exchange between water and steel, we can say that the heat lost by the knife, must be equal to the heat gained by the water.

As we have a limit for the maximum temperature of both elements (once reached a final thermal equilibrium), of 100ºC, which means that the maximum allowable change in temperature will be of 300º C for the knife, and of 80º C for the water.

Empirically , it has been showed that for a heat exchange process using only conduction, the heat needed to raise the temperature of a body, is proportional to the mass, being the proportionality constant a factor that depends on the material, called specific heat.

So, we can write the following equation:

cs*mk*Δtk = cw*mw*Δtw

Replacing by the givens of the question, we have:

0.11 cal/gºC * 80 g * 300ºC = 1 cal/gºC*mw*80ºC

Solving for mw = 2,640 cal / 80 cal/g =33 g.

measuring things 1 litre of water

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.