Suppose that you have 23 g of silver. Also assume that this sample of silver is initially at 18 oC. What is the final temperature of this sample of silver after it absorbs 2.05 kJ of heat? csilver = 0.235 J/goC Enter your answer numerically, in terms of oC.

396.92°C

Explanation:

Hello,

This question relates to heat transfer between objects and we can simply use the formula to solve this question.

Data;

Mass of silver (m) = 23g

Initial temperature of silver (T1) = 18°C

Final temperature of silver (T2) = ?

Heat absorbed = 2.05kJ = 2050J

Specific heat capacity of silver = 0.235J/g°C

Heat energy (Q) = mc∇T

Q = MC∇T

Q = heat energy

M = mass of substance

C = specific heat capacity of substance

∇T = change in temperature of substance = (T2 - T1)

Q = MC∇T

2050 = 23 * 0.235 * (T2 - 18)

2050 = 5.41 * (T2 - 18)

2050 = 5.41T2 - 97.38

Collect like terms

2050 + 97.38 = 5.41T2

2147.38 = 5.41T2

Divide both sides by 5.41

2147.38 / 5.41 = 5.41T2 / 5.41

T2 = 396.92°C

The final temperature of the sample after absorbing 2.05kJ of heat is 396.92°C

Have a nice day