How much heat is needed to bring 12.0 g of water from 28.3 °C to 43.87 °C, if the specific heat capacity of water is 4.184 J/(g•
1 answer:
Answer:
Heat capacity, Q = 781.74 Joules
Explanation:
Given the following data;
Mass = 12g
Initial temperature = 28.3°C
Final temperature = 43.87°C
Specific heat capacity of water = 4.184J/g°C
To find the quantity of heat needed?
Heat capacity is given by the formula;
Where;
Q represents the heat capacity or quantity of heat.
m represents the mass of an object.
c represents the specific heat capacity of water.
dt represents the change in temperature.
dt = T2 - T1
dt = 43.87 - 28.3
dt = 15.57°C
Substituting into the equation, we have;
Q = 781.74 Joules
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