There are two lead balls. How many times is the mass of the first ball greater than the mass of the second, if 3 times more heat is spent to heat the first ball by 1.5 K than to heat the second by 1 K?
The amount of heat is determined from the following expression:
Q = c * m * (t2-t1)
For the first ball (we have the same c, because the material is the same):
Q1 = c * m1 * (t2-t1) = c * m1 * 1.5
For the second:
Q2 = c * m2 * (t2-t1) = c * m ** 1
Let us express the masses from these expressions:
m1 = Q1 / 1.5 * s
m2 = Q2 / 1 * s
Let’s divide m1 / m2:
m1 / m2 = (Q1 / 1.5 * s) / (Q2 / 1 * s) = Q1 / (1.5 * Q2) = 2
Answer: 2 times.
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