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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Answer:
b) TA = TB = TC
Explanation:
- When put in contact each other, and isolated, both blocks will exchange heat till they reach to thermal equilibrium.
- During this process, the body at a higher temperature, will loss heat, tat it will be gained by the other body.
- The equilibrium condition will be reached when the following equation be met:
- Replacing by the values of T₀A = 300º C, and T₀B = 400ºC, and simplifying common terms as mA = mB, we can solve for Tfin, as follows:
- So, when both blocks reach to equilibrium, they will be at a common final temperature, 350ºC.
- When put in contact with block C, at the same temperature, at that instant, the three blocks will have the same common temperature of 350 ºC.
- So, option b) is the right one.