Answer:
THE INITIAL TEMPERATURE OF THE SECOND SAMPLE IS 4.93 C OR 277.93 K
Explanation:
Mass of first sample of water = 106 g
Initial temp of first sample = 21.4 °C = 21.4 + 273 K = 294.4 K
Mass of second sample = 64.3 g
Final temp of theresulting mixture = 46.8 °C = 46.8 + 273 K = 319.8 K
Specific heat capacity of water = 4.184 J/g K
It is worthy to note that;
Heat gained by the first sample = Heat lost by the second sample
Since heat = mass * specific heat capacity * change in temperature, we have
Mass * specific heat * change in temp of the first sample = Mass * specific heat * change in temp. of the second sample
MC (T2 - T1) = MC (T2-T1)
106 * 4.184 * ( 319.8 - 294.4) = 64.3 * 4.184 * ( 319.8 - T1)
106 * 4.184 * 25.4 = 269.0312 ( 319.8 - T1)
11 265.0016 = 269.0312 (319.8 - T1)
Since the change in temperature = 319.8 -T1
Change in temperature =11265.0016 / 269.0312
Change in temperature = 41.87
Change in temperature = 319.8 -T1
41.87 = 319.8 - T1
T1 = 319.8 - 41.87
T1 = 277.93 K
T1 = 4.93 °C
So therefore, the initial temperature of the sacond sample is 4.73 °C or 277.93 K
