Question

A silver piece of metal is thrown put into some water.

Information on water
Mass of water= 1 kg
Initial Temperature of water = 295 K
Final Temperature of water =303 K
Cwater=4180 J/kg K
How many Joules of heat are added to the water?

Answer 1

Q = 33440 Joules.

Step-by-step explanation:

Given the following data;

Mass of water= 1 kg

Initial Temperature of water = 295 K

Final Temperature of water =303 K

C of water=4180 J/kg K

To find the quantity of heat;

Heat capacity is given by the formula;

`Q = mcdt`

Where,

Q represents the heat capacity.

m represents the mass of an object.

c represents the specific heat capacity of water.

dt represents the change in temperature.

dt = T2 - T1

dt = 303 - 295

dt = 8K

Substituting into the equation, we have

Q = 1*4180*8

Q = 33440 Joules.