What mass of water at 10.0 oc would be required to cool 50.0g of a metal having a specific heat of
0.60 j/g. oc from 90.0 oc to 20.0 oc?

The mass of water required is 50.2 g

That is;

Q=m×c×ΔT

In this case, water is used to cool a metal, therefore, water will gain heat while the metal will lose heat.

Heat gained is equivalent to heat lost

Heat gained by water = Heat lost by the metal

Step 1: Heat lost by the metal

Mass of the metal  = 50.0 g

Specific heat capacity of metal = 0.60 J/g°C

Change in temperature (ΔT) = 70°C

Heat lost by the metal = 50 g× 0.6 × 70

                                     = 2100 Joules

Step 2: Heat gained by water

Mass of water = x g

Specific heat capacity of water = 4.18 J/g°C

Temperature change = 10 °C

Heat gained by water = x g × 4.18 × 10

                                    = 41.8x joules

Step 3: Mass of water

Heat gained by water = heat lost by the metal

41.8x = 2100

   x = 50.239 g

      = 50.2 g (1 d.p)

The mass of water is 50.2 g

answer: a tank of height h is filled with water and sits on the ground. water squirts from a hole at a height y above the ground and has a range r. for two y values, 0 and h, r is zero. determine the value of y for which the range will be a maximum.

explanation:

it can be observed without changing the identity of the object.

it depends on what substance you're talking about as well as many other factors. in quantum mechanics, the observation of particles affects the observers ability to measure certain properties. for example: you can know the velocity of a particle, but you can't know it's position; or you can know it's position, but not it's velocity. it's like trying to find out where a grain of sand is by throwing other grains of sand at it. once you find out where it is, it's been moved. it's also like trying to find an answer to the question of whether or not a tree falling in a forest makes a sound if nobody is there to hear it. paradoxically, the only way to find out is to be there and hear it.