Answer:
Heat transfer = Q = 62341.6 J
Explanation:
Given data:
Heat transfer = ?
Mass of water = 50.0 g
Initial temperature = 30.0°C
Final temperature = 55.0°C
Specific heat capacity of water = 4.184 J/g.K
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 55.0°C - 30.0°C
ΔT = 25°C (25+273= 298 K)
Q = 50.0 g × 4.184 J/g.K ×298 K
Q = 62341.6 J
Taking into account the definition of calorimetry, the heat transfered is 5225 J.
Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.
Sensible heat is the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).
The expression for sensible heat in an isobaric process, that is, at constant pressure, is:
Q = c× m× ΔT
where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.
In this case, you know:
Replacing:
Q= 4.18 [tex]\frac{J}{gK}[/tex]× 50 g× 25 K
Solving:
Q=5225 J
Finally, the heat transfered is 5225 J.
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