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

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

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  

Q water = Q metal

plants need sunlight, water, and carbon dioxide to make food. plants use the raw materials of water, nutrients they absorb from soil, and carbon dioxide to make sugars the plant can use. energy from the sun and chlorophyll are also needed.