Question

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?

Answer 1

`"8.74 J/g K"`

Explanation:

Specific heat is given by

`"S" = "Q"/"mΔT"`

Where

• `"Q ="` Heat added/liberated
• `"m ="` Mass of sample
• `"ΔT ="` Change in temperature

`"S" = "3190 J"/"400 g × (308 K – 273 K)" = "8.74 J/g K"`