Answer:
[tex]Q=80cal[/tex]
Explanation:
Hello,
In this case, the relationship between heat, mass, specific heat and change in temperature is understood by:
[tex]Q=mCp(T_2-T_1)[/tex]
In this case, since the involved substance is water, whose specific heat is 1 cal/(g°C), we compute the heat 4 mL of water need to rise the temperature from 10 °C to 30 °C as shown below:
[tex]Q=4mL*\frac{1g}{1mL}*1\frac{cal}{g\°C}*(30\°C-10\°C)\\ \\Q=80cal[/tex]
Best regards.