3. how is temperature affected when? (a) the depth increases (b)the latitude increases

answer a or b

(A) In case of a water body such as oceans and seas, the temperature decreases with the increasing depth. It is because at higher depths the sunlight is not able to penetrate as a result of which the lower portion of the oceans remains dark and maintains a much colder temperature. But near the surface water layer, sunlight is penetrable so it remains relatively warmer.

In case of subsurface condition i.e. below the earth's surface, the temperature is directly proportional to the depth. The temperature is maximum at the core which is comprised of hot molten materials such as iron, nickel and other siderophile elements. This increase in the temperature with increasing depth is known as the geothermal gradient.

(B) The latitude directly controls the temperature of a particular area. The equatorial region receives maximum amount of sunlight so they experiences relatively very high temperature. As one move towards the poles the temperature gradually drops down and is the minimum at the poles.

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol