How much heat would be required to heat 1 gram of pure liquid water from 10C to 20C?

How would i approach this problem?

20-10=10*1*4.18=41.8 joul/1 gm

1 cal = 4.18 joul
s.h.w=1 cal/gm

To approach this problem, you can use the formula for specific heat capacity. The specific heat capacity of water is approximately 4.18 Joules per gram per degree Celsius (J/g°C). Here are the steps to calculate the amount of heat required:

Step 1: Find the temperature change
The temperature change is the difference between the final temperature (20°C) and the initial temperature (10°C):
Temperature change = Final temperature - Initial temperature = 20°C - 10°C = 10°C

Step 2: Determine the mass of water
Given that you have 1 gram of water, the mass of water is 1 gram.

Step 3: Calculate the amount of heat required
To calculate the amount of heat required, use the formula:
Q = mass × specific heat capacity × temperature change
Q = 1 gram × 4.18 J/g°C × 10°C
Q = 41.8 Joules

Therefore, the amount of heat required to heat 1 gram of pure liquid water from 10°C to 20°C is 41.8 Joules.

To calculate the amount of heat required to heat a substance, you can use the specific heat formula. The specific heat capacity of a substance is the amount of heat energy required to raise the temperature of 1 gram of that substance by 1 degree Celsius.

In this case, we need to find the amount of heat required to heat 1 gram of pure liquid water from 10°C to 20°C.

The specific heat capacity of water is 4.184 J/g°C. This means that it takes 4.184 joules of energy to raise the temperature of 1 gram of water by 1 degree Celsius.

To calculate the amount of heat required, you can use the following formula:

Q = m * c * ΔT

Where:
Q - Quantity of heat energy (in joules)
m - Mass of the substance (in grams)
c - Specific heat capacity of the substance (in J/g°C)
ΔT - Change in temperature (in °C)

For this problem, the mass (m) is 1 gram, the specific heat capacity (c) is 4.184 J/g°C, and the change in temperature (ΔT) is 20°C - 10°C = 10°C.

Plugging these values into the formula, we get:

Q = 1g * 4.184 J/g°C * 10°C
Q = 41.84 joules

So, it would require 41.84 joules of heat energy to raise the temperature of 1 gram of pure liquid water from 10°C to 20°C.

figured it out.