A 150 gram golden (specific heat 0.03 cal/gram^0 C) spoon at 20^0C is placed in a cup of tea at 100^0C. The amount of heat which the spoon absorbs from the tea which is necessary to reach spoon's tempature 47^0C is

a)360.4 Cal b)121.5 Cal c)78.24 Cal d)15.08 Cal e)157.5 Cal

To find the amount of heat the spoon absorbs from the tea to reach a temperature of 47°C, we can use the equation:

Q = mcΔT

Where:
Q = amount of heat transferred
m = mass of the spoon (in grams)
c = specific heat of the material (in cal/gram°C)
ΔT = change in temperature (final temperature - initial temperature)

Given:
m = 150 grams
c = 0.03 cal/gram°C
Initial temperature of the spoon (T1) = 20°C
Final temperature of the spoon (T2) = 47°C
Tea temperature (Ttea) = 100°C

First, we need to calculate the amount of heat transferred from the tea to the spoon until the spoon reaches the temperature of the tea:

Q1 = mcΔT1

ΔT1 = Ttea - T1
= 100°C - 20°C
= 80°C

Q1 = 150 grams × 0.03 cal/gram°C × 80°C
= 360 cal

Next, we need to calculate the amount of heat required for the spoon to reach the final temperature of 47°C from its initial temperature of 20°C:

Q2 = mcΔT2

ΔT2 = T2 - T1
= 47°C - 20°C
= 27°C

Q2 = 150 grams × 0.03 cal/gram°C × 27°C
= 121.5 cal

Lastly, we need to sum up the heat transferred in Q1 and Q2:

Total heat absorbed by the spoon = Q1 + Q2
= 360 cal + 121.5 cal
= 481.5 cal

Converting cal to Cal (calorie to Calorie):
1 cal = 0.001 Cal

Total heat absorbed in Calories = 481.5 cal × 0.001 Cal/cal
= 0.4815 Cal

Therefore, the amount of heat the spoon absorbs from the tea to reach a temperature of 47°C is approximately 0.4815 Cal. None of the given answer options perfectly match this value, but the closest option is 0.4815 Cal, which corresponds to none of the given answer options.