Question

How many calories of heat are required to raise the temperature of 525g of

Aluminum from 13.0°C to 47.8°C? (CAL= 0.21 cal/g°C)

Answer 1

Step-by-step explanation:

Hi there,

To get started, recall the Heat-specific heat capacity equation of a substance:

Q=mCΔT

Our final temperature is 47.8 °C, since this is the way it is worded in the response (from temperature X to temperature Y)

Quite simply, we can go ahead and plug in mass, specific heat capacity, and change in temperature as all the units match up!

`Q= (525 g)(0.21 \ cal\ / g*C^(o) )(47.8C^(o)-13.0 C^(o))=3836.7 \ cal`

Study well and persevere.

thanks,