# It takes 3,190 J to increase the temperature of a 0.400 kg sample of glass from 273 K to 308 K. What is the specific heat for this type of glass?

May 29, 2018

$\text{8.74 J/g K}$

#### Explanation:

Specific heat is given by

$\text{S" = "Q"/"mΔT}$

Where

• $\text{Q =}$ Heat added/liberated
• $\text{m =}$ Mass of sample
• $\text{ΔT =}$ Change in temperature

$\text{S" = "3190 J"/"400 g × (308 K – 273 K)" = "8.74 J/g K}$