Question

Q = cmAT

Answer 1

Answer: $Q=3000 cal$

Explanation:

We are given the following formula:

$Q=m. c. \Delta T$   (1)

Where:

$Q=3000 cal$ is the amount of heat

$m=300g$  is the mass  of water

$c=1 cal/g \°C$  is the specific heat of water

$\Delta T$  is the variation in temperature, which in this case is  $\Delta T=30\°C-20\°C=10\°C$  

Rewriting equation (1) with the known values at the right side, we will prove the result is $3000 cal$:

$Q=(300g)(1 cal/g \°C)(10\°C)$   (2)

$Q=3000 cal$   This is the result