B. Hot water The specific heat capacity of water is 4.186J/g °C
Calculate the temperature change when 8000 j of heat is added to 3g of water.
Use the equation q=mcAT
C. Specific heat capacity
Water has a specific heat capacity of 4.186J/g °C
Concrete has a specific heat capacity of 0.880 J/g °C
If a kg of water and a kg of concrete sit out in the sun on a hot day which one will have a change of temperature faster?

Explain why?

The heat capacity measures the amount of heat necessary to raise the temperature of an object or system by one degree Celsius.

LEARNING OBJECTIVES

Explain the enthalpy in a system with constant volume and pressure

KEY TAKEAWAYS

Key Points

Heat capacity is the measurable physical quantity that characterizes the amount of heat required to change a substance’s temperature by a given amount. It is measured in joules per Kelvin and given by.

The heat capacity is an extensive property, scaling with the size of the system.

The heat capacity of most systems is not constant (though it can often be treated as such). It depends on the temperature, pressure, and volume of the system under consideration.

Key Terms

a chemical reaction

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