Question

Please help!!!
When 38 g of a metal at 92 °C is added to
43 g of water at 22°C, the temperature of the
water rises to 30°C. What is the specific heat
capacity of the metal? Assume no heat was
lost to the surroundings.

Answer in units of
J/g degrees C

Answer 1

The specific heat  capacity of the metal : 0.610 J/g° C  

Further explanation  

The law of conservation of energy can be applied to heat changes, i.e. the heat received/absorbed is the same as the heat released  

Q abs = Q received  

Heat can be calculated using the formula:  

Q = mc∆T  

Q = heat, J  

m = mass, g  

c = specific heat, joules / g ° C  

∆T = temperature difference, ° C / K  

∆T = T(final temp) - T(initial temp)

mass of metal = 38 g

T initial metal = 92 °C

mass of water = 43 g

T final water = 30  °C

T initial water = 22°C

c water = 4.18 J/g° C  

• Heat absorbed by water

$\tt Q=m.c.\Delta T\\\\Q=43\times 4.18\times (30-22^oC)\\\\Q=1437.92~J$

• the specific heat  capacity of the metal

Q water = Q metal

$\tt 1437.92=38\times c\times (92-30)\\\\c=0.610~J/g^oC$