Question

How many kilojoules of heat must be applied to warm 90 g of water from 25°C to 57°C? (Note: The specific heat capacity of liquid water is 4.184J/g ∘ C)

a.1060.56 kJ
b.25.11 kJ
c.29.48 kJ
d.31.98 kJ

Answer 1

Approximately 11.996 kJ of heat must be applied to warm 90 g of water from 25°C to 57°C.

Step-by-step explanation:

To calculate the amount of heat required to warm the water, we can use the equation:

Q = mcΔT

where Q is the heat energy, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.

Given:

• Mass of water (m) = 90 g
• Specific heat capacity of water (c) = 4.184 J/g °C
• Change in temperature (ΔT) = (57 °C - 25 °C) = 32 °C

Substituting these values into the equation:

Q = (90 g)(4.184 J/g °C)(32 °C) = 11995.52 J = 11.996 kJ

Therefore, approximately 11.996 kJ of heat must be applied to warm 90 g of water from 25°C to 57°C.