How much heat in joules is required to heat a 50g sample of aluminum from 73°F to 148°F?

Question

Express answer using two significant figures.

Answer 1

What we need to do is incorporate the following equation. Amount of heat necessary=specific heat capacity*grams*change in temperature.

The specific heat capacity for aluminum is 0.89 J/ g degrees Celsius.

We need to convert the temperatures to Celsius.

T_(°C) = (T_(°F) - 32) × 5/9

(73-32)*5/9=22.78 degrees Celsius (Temperature corresponding to 73 degrees Fahrenheit

We do the same for 148 degrees Fahrenheit: (148-32)*5/9=64.44.

Now, let's calculate: 0.89*(64.44-22.78)*50 grams=1853.87 J needed to heat the sample