A 113.25 gram sample of gold is initially at 100.0 ^°C. It gains 20.00 J of heat from its surroundings. What is its final temperature? (specific heat of gold = 0.129 J g^-1 ^°C^-1)

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The heat gained by the gold is given by:Q = mC_pΔTwhere:Q is the heat gained (in J)m is the mass of the gold (in g)C_p is the specific heat of gold (in J g^-1 ^°C^-1)ΔT is the change in temperature (in ^°C)Substituting the given values into the equation, we get:20.00 J = (113.25 g)(0.129 J g^-1 ^°C^-1)ΔTSolving for ΔT, we get:ΔT = 20.00 J / (113.25 g)(0.129 J g^-1 ^°C^-1) = 1.43 ^°CTherefore, the final temperature of the gold is:T_final = T_initial + ΔT = 100.0 ^°C + 1.43 ^°C = 101.4 ^°C

A 11325 Gram Sample of Gold Is Initially at 100