temperature

Question

AS cube of gold weighting 192.4 grams is heated from 30.0 degrees C to some higher temperature, with the absorption of 226 calories of heat. The specific heat of gold is 0.0300 cal/g degrees C. What is was the final temperature of the gold? 

 

Answer 1

Use the relation

 

Q = mc(T_f - T_i)

 

T_f = T_i + Q/(mc) = 30.0°C + (226 cal) / [(192.4 g)(0.0300 cal/ g°C) = 69.15°C