How many grams of water would require 2.20 x 10 4joules of heat to raise its temperature from 34.0°C to 100.0°C? The specific heat of water is 4.18 J/g∙°C
To determine the grams of water that would require 2.20 x 10^4 joules of heat, we need to calculate the amount of heat absorbed by the water during the temperature change.
The equation to calculate the amount of heat absorbed is:
q = m * c * ΔT
where:
q = amount of heat absorbed
m = mass of the water (grams) - unknown
c = specific heat of water (4.18 J/g∙°C)
ΔT = change in temperature (final temperature - initial temperature)
Substituting the given values into the equation:
2.20 x 10^4 J = m * (4.18 J/g∙°C) * (100.0°C - 34.0°C)
Simplifying:
2.20 x 10^4 J = m * 4.18 J/g∙°C * 66.0°C
Dividing both sides of the equation by (4.18 J/g∙°C * 66.0°C):
2.20 x 10^4 J / (4.18 J/g∙°C * 66.0°C) = m
Simplifying:
m = 79.686 g
Therefore, it would require 79.686 grams of water to raise its temperature from 34.0°C to 100.0°C.
To solve this problem, we can use the formula:
Q = mcΔT
where:
Q is the amount of heat energy absorbed or released,
m is the mass of the substance,
c is the specific heat capacity of the substance, and
ΔT is the change in temperature.
Given:
Q = 2.20 × 10^4 J
ΔT = (100.0 - 34.0) °C = 66.0 °C
c (specific heat capacity) = 4.18 J/g·°C
Let's solve for the mass of water (m):
Q = mcΔT
Rearranging the formula:
m = Q / (cΔT)
Substituting the given values:
m = (2.20 × 10^4 J) / (4.18 J/g·°C × 66.0 °C)
m = 2.20 × 10^4 J / (275.88 J/g)
Calculating:
m ≈ 79.67 g
Therefore, approximately 79.67 grams of water would require 2.20 × 10^4 joules of heat to raise its temperature from 34.0°C to 100.0°C.
To find the number of grams of water required to raise its temperature, you can use the formula:
Q = m * c * ΔT
Where:
Q = heat energy in joules
m = mass of water in grams
c = specific heat of water in J/g°C
ΔT = change in temperature in °C
In this case, we are given:
Q = 2.20 x 10^4 J
c = 4.18 J/g∙°C
ΔT = (100.0°C - 34.0°C) = 66.0°C
Therefore, the equation can be rearranged to solve for 'm' (mass of water):
m = Q / (c * ΔT)
Substituting the given values:
m = 2.20 x 10^4 J / (4.18 J/g∙°C * 66.0°C)
Now, we can calculate:
m = 320.57 g
Therefore, approximately 320.57 grams of water would require 2.20 x 10^4 joules of heat to raise its temperature from 34.0°C to 100.0°C.