If 30.45 grams of water is to be heated up 3.3 degrees to make baby
1 answer:
Answer:
A. 420 J
Explanation:
Given the following data;
Mass = 30.45 g
Specific heat capacity = 4.18 J/g °C.
Temperature = 3.3°C
To find the quantity of heat;
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.
t represents the temperature.
Substituting into the equation, we have;
Q = 420.03 ≈ 420 Joules
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Answer:
The final velocity of the runner at the end of the given time is 2.7 m/s.
Explanation:
Given;
initial velocity of the runner, u = 1.1 m/s
constant acceleration, a = 0.8 m/s²
time of motion, t = 2.0 s
The velocity of the runner at the end of the given time is calculate as;
where;
v is the final velocity of the runner at the end of the given time;
v = 1.1 + (0.8)(2)
v = 2.7 m/s
Therefore, the final velocity of the runner at the end of the given time is 2.7 m/s.