Two copper balls of masses m0 and 4m0 are heated so that both balls receive the same amount of heat.

Two copper balls of masses m0 and 4m0 are heated so that both balls receive the same amount of heat. At the same time, the large ball warmed up by 5 ° C. How hot was the smaller ball? How much heat is released when 4 m3 of ice is cooled from 10 ° C to –40 ° C?

Given:
Q1 = Q2
m1 = m0
m2 = 4m0
t2 = 4
t1 =?
Decision:
Q = cm [t]
Q1 = cm0t
Q2 = c4m0 * 5
Q1 / Q2 = cmt1 / c4m * 5
t1 = 4 * 5 = 20
Answer: a ball with a smaller mass warmed up by 20 degrees Celsius