How much heat is released when 345 g of tin is cooled from 210.0 degrees * C to 0.0 C? (C= 0.220 J/gC)

18634 joules

Explanation:

The quantity of Heat Energy (Q) released when a heated substance is cooled depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Since,

Q = ?

Mass of oil = 345g

C = 0.220 J/g°C

Φ = (Final temperature - Initial temperature)

= 0°C - 220°C = -220°C

Then, Q = MCΦ

Q = 345g x 0.220 J/g°C x -220°C

Q = 18634 joules (note that negative sign will not reflect in answer, because it only indicates that heat was released rather than gained)

Thus, 18634 joules of heat is released

c- cold periods in earth's history during which glaciers covered much of the surface.

explanation:

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