Answer : Energy needed, Q = 0.3 cal
Explanation :
Given that,
The specific heat capacity of silver, c = 0.06 cal/g °C
Mass of the sample, m = 5 grams
It changes its temperature from 10°C to 11°C, \(\Delta T=1^oC\)
Energy need to flow is given by :
\(Q=mc\Delta T\)
\(Q=5\ g\times 0.06\ cal/g\ ^oC\times 1^oC\)
\(Q=0.3\ cal\)
So, the energy needed to flow into this sample is 0.3 cal
Hence, this is the required solution.
