Question

How much heat is required to warm 230 g of water from 12°C to 90°C (Specific Heat of Water = 4.184)?

A) 82,958.4 J
B) 96,176.8 J
C) 88,419.2 J
D) 74,201.6 J

Answer 1

The heat required to warm 230 g of water from 12°C to 90°C with a specific heat capacity of 4.184 J/g°C is 74,201.6 joules (Option D).

Step-by-step explanation:

To calculate the amount of heat required to warm a given mass of water from one temperature to another, we use the formula:

q = mcΔT

where:

• q is the heat energy (in joules, J)
• m is the mass of water (in grams, g)
• c is the specific heat capacity (in J/g°C)
• ΔT is the change in temperature (in °C)

Given:

• The mass of the water, m, is 230 g.
• The initial temperature is 12°C, and the final temperature is 90°C, thus ΔT = 90°C - 12°C = 78°C.
• The specific heat of water is 4.184 J/g°C.

Substitute these values into the formula:

q = (230 g)(4.184 J/g°C)(78°C)

q = 230 g * 4.184 J/g°C * 78°C

q = 74,201.6 J

Thus, the heat required is 74,201.6 joules, which corresponds to Option D.