Apan containing 30 grams of water was allowed to cool from a temperature of 90.0 °c. if the amount of heat released is 1,500 joules, what is the approximate final temperature of the water? a. 76 °c b. 78 °c c. 81 °c d. 82 °c

B. 78 °C.

Explanation:

Q = m.c.ΔT,

where, Q is the amount of heat released from water (Q = - 1500 J).

m is the mass of water (m = 30.0 g).

c is the specific heat capacity of water (c = 4.18 J/g.°C).

ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = final T - 90.0 °C).

∴ (- 1500 J) = (30.0 g)(4.18 J/g.°C)(final T - 90.0 °C)

∴ (final T - 90.0 °C) = (- 1500 J)/(30.0 g)(4.18 J/g.°C) = - 11.96°C.

∴ final T = 90.0 °C - 11.96°C = 78.04°C ≅ 78 °C.

So, the right choice is: B. 78 °C.

h let me think

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