Respuesta :
What is the amount of heat, in joules, required to increase the temperature of a 49.5-gram sample of water from 22°C to 66°C?
- (1) 9100 J
- (2) 4600 J
- (3) 1400 J
- (4) 2300 J
Answer:
9100 J is the amount of heat, in joules, required to increase the temperature of a 49.5-gram sample of water from 22°C to 66°C.
Explanation:
Given: Q = ? J
mass = 49.5 g = 0.0495 kg
T1 = 22°C
T2 = 66°C
Cp = 4.18 J/Kg°C
Q = mCpΔT
Q = (4.95)(4.18)(66 - 22)
Q = 9104 J
So 9100 J is the amount of heat, in joules, required to increase the temperature of a 49.5-gram sample of water from 22°C to 66°C.
The heat required to raise the temperature of the water sample has been 910.40 J.
The specific heat has been defined as the amount of heat required to raise the temperature of 1 gram of a substance by 1 degree Celsius.
The specific heat has been expressed as:
[tex]\text {Heat} = mc\Delta T[/tex]
Where, the mass of water sample, [tex]m=4.95\;\text g[/tex]
The specific heat of water, [tex]c=4.18\;\rm{J/g^\circ C}[/tex]
The change in temperature of the water,[tex]\Delta T=T_i\;-\;T_f[/tex]
The value of change in temperature has been given by:
The initial temperature of the system,[tex]T_i=22^\circ \text C[/tex]
The final temperature of the system, [tex]T_f=66^\circ \text C[/tex]
Substituting the values for change in temperature ([tex]\Delta T[/tex]),
[tex]\Delta T=66-22\;^\circ \text C\\\Delta T=44\;^\circ \text C[/tex]
The change in temperature of the water, [tex]\Delta T=44\;^\circ \text C[/tex]
Substituting the values for the heat required,
[tex]\rm Heat=4.95\;\times\;4.185\;\times\;44\;\text J\\Heat=910.40\;\text J[/tex]
The heat required to raise the temperature of the water sample has been 910.40 J.
For more information about specific heat, refer to the link:
https://brainly.com/question/8647002
