How much heat is required to raise the temperature of 20.0g of H2O from 18.0 Celcius to 28.0 Celcius?
heat=mass*specificheatcontent*deltaTemp
We have,
E = (mass)(specific heat capacity)(delta Temp)
mass = 20.0/1000 = 0.02 kg
=> E = (0.0200kg)(4200J/kg C)(28.0C-18.0C)
= 840 J
To find the amount of heat required to raise the temperature of a substance, we can use the formula:
q = m * c * ΔT
where:
q = heat energy
m = mass of the substance
c = specific heat capacity
ΔT = change in temperature
Let's break down the calculation step by step:
1. Find the change in temperature (ΔT):
ΔT = final temperature - initial temperature
= 28.0 °C - 18.0 °C
= 10.0 °C
2. Determine the specific heat capacity of water (c):
The specific heat capacity of water is approximately 4.18 J/g·°C.
3. Calculate the heat energy (q):
q = m * c * ΔT
= 20.0 g * 4.18 J/g·°C * 10.0 °C
Now, we can calculate the result:
q = 8360 J
Therefore, it would require 8360 Joules (J) of heat energy to raise the temperature of 20.0 grams of water from 18.0 °C to 28.0 °C.