Problem A
E = m * c * delta T
E = 6.30 * 0.377 * (32.0 - 20.0)
E = 6.30 * 0.377 * 12
E = 28.5012 Now you are asked for three sig digs.
E = 28.5 Joules is your answer.
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Problem B
Energy gained by water = Energy lost by limestone
Let the initial temp of the limestone = t
Water
E_water = m*c * deltat_water
E_water = 75 * 4.186 * (51.9 - 23.1)
E_water = 75 * 4.186 * 28.8
E_water = 9041.76
Limestone
E_limestone = 62*0.921(T - 51.9)
E_limestone = 57.6546 (T - 51.9)
E_limestone = 57.6546*T - 57.6546 * 51.9
E_limestone= 57.6546*T - 2992.27
Now equate the two parts.
9041.76 = 57.6546T - 2992.27 Add 2992.27 to both sides.
9041.76 + 2992.27 = 57.6546T Combine
12034.03 = 57.654T Divide by 57.654
12034.03 / 57.654 =T Divide and switch.
T = 208.72
If you are to present this in 3 sig digs then the answer is 209oCelcius.