Question

The temperature of a 95.4 g piece of metal increases from 15° to 38° When the metal absorbs 849J of heat what is the specific heat up the metal

Answer 1

A substance's specific heat tells you how much heat much be provided to increase the temperature of 1g of the substance by $1^{0}C$

$q=m×c×\Delta T$ Thus $c=\frac{q} {m×\Delta T}$

$c=\frac{849J}{95.4g×(48-25) ^{0}C}$

$c=0.3869J/g^{0}C$