Question

It takes 2,267 joules of heat to raise the temperature of a 44.5 gram sample of a metal from 33.9°C to 288.3°C. What is the heat capacity of the metal?

Answer 1

Answer:

$\boxed{\text{0.200 J ^{\circ} C ^{-1} g ^{-1} }}}$

Explanation:

The formula for the amount heat q absorbed by a substance is

q = mcΔT

where

 m = the mass of the substance

  C = the specific heat capacity of the material

ΔT = the temperature change

Data:

 q = 2267 J

m = 44.5 g

T₁ = 33.9 °C

T₂ = 288.3 °C

Calculations:

ΔT = (288.3 - 33.9) °C = 254.4 °C

$\begin{array}{rcl}2267 & = & 44.5 \times C \times 254.4\\2267 & = & 11 180C\\C& = & \dfrac{2267}{11180}\\\\& = & \textbf{0.200 J ^{\circ} C ^{-1} g ^{-1} }\\\end{array}$

$\text{The specific heat capacity of the metal is \boxed{\textbf{0.200 J ^{\circ} C ^{-1} g ^{-1} }}}$